63
Determine the common difference, and find the next four terms of each arithmetic sequence. 1. 20, 17, 14, SOLUTION: First, find the common difference. 17 20 = 3 14 17 = 3 The common difference is 3. Add 3 to the third term to find the fourth term, and so on. 14 + (3) = 11 11 + (3) = 8 8 + (3) = 5 5 + (3) = 2 Therefore, the next four terms are 11, 8, 5, and 2. 2. 3, 16, 29, SOLUTION: First, find the common difference. 16 3 = 13 29 16 = 13 The common difference is 13. Add 13 to the third term to find the fourth term, and so on. 29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81 Therefore, the next four terms are 42, 55, 68, and 81. 3. 117, 108, 99, SOLUTION: First, find the common difference. 108 117 = 9 99 108 = 9 The common difference is 9. Add 9 to the third term to find the fourth term, and so on. 99 9 = 90 90 9 = 81 81 9 = 72 72 9 = 63 Therefore, the next four terms are 90, 81, 72, and 63. 4. eSolutions Manual - Powered by Cognero Page 1 10-2 Arithmetic Sequences and Series

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Page 1: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 1

10-2 Arithmetic Sequences and Series

Page 2: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 2

10-2 Arithmetic Sequences and Series

Page 3: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 3

10-2 Arithmetic Sequences and Series

Page 4: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 4

10-2 Arithmetic Sequences and Series

Page 5: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 5

10-2 Arithmetic Sequences and Series

Page 6: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 6

10-2 Arithmetic Sequences and Series

Page 7: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 7

10-2 Arithmetic Sequences and Series

Page 8: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 8

10-2 Arithmetic Sequences and Series

Page 9: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 9

10-2 Arithmetic Sequences and Series

Page 10: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 10

10-2 Arithmetic Sequences and Series

Page 11: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 11

10-2 Arithmetic Sequences and Series

Page 12: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 12

10-2 Arithmetic Sequences and Series

Page 13: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 13

10-2 Arithmetic Sequences and Series

Page 14: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 14

10-2 Arithmetic Sequences and Series

Page 15: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 15

10-2 Arithmetic Sequences and Series

Page 16: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 16

10-2 Arithmetic Sequences and Series

Page 17: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 17

10-2 Arithmetic Sequences and Series

Page 18: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 18

10-2 Arithmetic Sequences and Series

Page 19: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 19

10-2 Arithmetic Sequences and Series

Page 20: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 20

10-2 Arithmetic Sequences and Series

Page 21: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 21

10-2 Arithmetic Sequences and Series

Page 22: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 22

10-2 Arithmetic Sequences and Series

Page 23: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 23

10-2 Arithmetic Sequences and Series

Page 24: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 24

10-2 Arithmetic Sequences and Series

Page 25: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 25

10-2 Arithmetic Sequences and Series

Page 26: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 26

10-2 Arithmetic Sequences and Series

Page 27: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 27

10-2 Arithmetic Sequences and Series

Page 28: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 28

10-2 Arithmetic Sequences and Series

Page 29: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 29

10-2 Arithmetic Sequences and Series

Page 30: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 30

10-2 Arithmetic Sequences and Series

Page 31: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 31

10-2 Arithmetic Sequences and Series

Page 32: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 32

10-2 Arithmetic Sequences and Series

Page 33: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 33

10-2 Arithmetic Sequences and Series

Page 34: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 34

10-2 Arithmetic Sequences and Series

Page 35: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 35

10-2 Arithmetic Sequences and Series

Page 36: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 36

10-2 Arithmetic Sequences and Series

Page 37: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 37

10-2 Arithmetic Sequences and Series

Page 38: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 38

10-2 Arithmetic Sequences and Series

Page 39: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 39

10-2 Arithmetic Sequences and Series

Page 40: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 40

10-2 Arithmetic Sequences and Series

Page 41: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 41

10-2 Arithmetic Sequences and Series

Page 42: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 42

10-2 Arithmetic Sequences and Series

Page 43: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 43

10-2 Arithmetic Sequences and Series

Page 44: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 44

10-2 Arithmetic Sequences and Series

Page 45: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

 

eSolutions Manual - Powered by Cognero Page 45

10-2 Arithmetic Sequences and Series

Page 46: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

Determine the common difference, and find the next four terms of each arithmetic sequence.1. 20, 17, 14, …

SOLUTION:  First, find the common difference. 17 – 20 = –3 14 – 17 = –3   The common difference is –3. Add –3 to the third term to find the fourth term, and so on.14 + (–3) = 11 11 + (–3) = 8 8 + (–3) = 5 5 + (–3) = 2   Therefore, the next four terms are 11, 8, 5, and 2.

2. 3, 16, 29, …

SOLUTION:  First, find the common difference. 16 – 3 = 13 29 – 16 = 13   The common difference is 13. Add 13 to the third term to find the fourth term, and so on.29 + 13 = 42 42 + 13 = 55 55 + 13 = 68 68 + 13 = 81   Therefore, the next four terms are 42, 55, 68, and 81.

3. 117, 108, 99, …

SOLUTION:  First, find the common difference. 108 –117 = –9 99 – 108 = –9   The common difference is –9. Add –9 to the third term to find the fourth term, and so on.99 – 9 = 90 90 – 9 = 81 81 – 9 = 72 72 – 9 = 63   Therefore, the next four terms are 90, 81, 72, and 63.

4. −83, −61, −39, …

SOLUTION:  First, find the common difference. –83 – 61 = 22 –61 – 39 = 22   The common difference is 22. Add 22 to the third term to find the fourth term, and so on.–39 + 22 = –17 –17 + 22 = 5 5 + 22 = 27 27 + 22 = 49   Therefore, the next four terms are –17, 5, 27, and 49.

5. −3, 1, 5, …

SOLUTION:  First, find the common difference. 1 – (–3) = 4 5 – 1 = 4   The common difference is 4. Add 4 to the third term to find the fourth term, and so on.5 + 4 = 9 9 + 4 = 13 13 + 4 = 17 17 + 4 = 21   Therefore, the next four terms are 9, 13, 17, and 21

6. 4, 21, 38, …

SOLUTION:  First, find the common difference. 21 – 4 = 17 38 – 21 = 17   The common difference is 17. Add 17 to the third term to find the fourth term, and so on.38 + 17 = 55 55 + 17 = 72 72 + 17 = 89 89 + 17 =106   Therefore, the next four terms are 55, 72, 89, and 106

7. −4.5, −9.5, −14.5, …

SOLUTION:  First, find the common difference. –9.5 –(–4.5) = –5 –14.5 – (–9.5) = –5   The common difference is –5. Add –5 to the third term to find the fourth term, and so on.–14.5 – 5 = –19.5 –19.5 – 5 = –24.5 –24.5 – 5 = –29.5 –29.5 – 5 = –34.5   Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.

8. −97, −29, 39, …

SOLUTION:  First, find the common difference. –29 – (–97) = 68 39 – (–29) = 68   The common difference is 68. Add 68 to the third term to find the fourth term, and so on.39 + 68 = 107 107 + 68 = 175 175 + 68 = 243 243 + 68 = 311    

9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band member, the second row has 3 band members, the third row has 5 band members, and so on. a. Find the number of band members in the 8th row. b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.

SOLUTION:  a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,

so a1 = 1, a2 = 3, and a3 = 5. Find the common difference.

3 – 1 = 2 5 – 3 = 2   The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term. 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 11 + 2 = 13 13 + 2 = 15   So, there are 15 band members in the 8th row.   b. For an explicit formula, substitute a1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 2

Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.

10. 2, 5, 8, …

SOLUTION:  First, find the common difference. 5 – 2 = 3 8 – 5 = 3   For an explicit formula, substitute a1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 2, an = an – 1 + 3

11. −6, 5, 16, …

SOLUTION:  First, find the common difference. 5 – (–6) = 11 16 – 5 = 11   For an explicit formula, substitute a1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –6, an = an – 1 + 11

12. −9, −16, −23, …

SOLUTION:  First, find the common difference. –16 – (–9) = –7 –23 – (–16) = –7   For an explicit formula, substitute a1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –9, an = an – 1 – 7

13. 4, 19, 34, …

SOLUTION:  First, find the common difference. 19 – 4 =15 34 – 19 = 15   For an explicit formula, substitute a1 = 4 and d = 15 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 4, an = an – 1 + 15

14. 25, 11, −3, …

SOLUTION:  First, find the common difference. 11 – 25 = –14 –3 – 11 = –14   For an explicit formula, substitute a1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 25, an = an – 1 – 14

15. 7, −3.5, −14, …

SOLUTION:  First, find the common difference. –3.5 – 7 = –10.5 –14 – (–3.5) = –10.5   For an explicit formula, substitute a1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 7, an = an – 1 – 10.5

16. −18, 4, 26, …

SOLUTION:  First, find the common difference. 4 – (–18 ) = 22 26 – 4 = 22   For an explicit formula, substitute a1 = –18 and d = 22 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= –18, an = an – 1 + 22

17. 1, 37, 73, …

SOLUTION:  First, find the common difference. 37 – 1 = 36 73 – 37 = 36   For an explicit formula, substitute a1 = 1 and d = 36 in the formula for the nth term of an arithmetic sequence.

  For the recursive formula, state the first term a1 and then indicate that the next term is the sum of the previous

term an – 1 and d.

  a1

= 1, an = an – 1 + 36

Find the specified value for the arithmetic sequence with the given characteristics.18. If a14 = 85 and d = 9, find a1.

SOLUTION:  

Substitute an = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.

 

19. Find d for −24, −31, −38, …

SOLUTION:  Find the difference between two pairs of consecutive terms.–31 – (–24) = –7 –38 – (–31) = –7   Therefore, d = –7.

20. If an = 14, a1 = −36, and d = 5, find n.

SOLUTION:  

Substitute an = 14, a1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.

21. If a1 = 47 and d = −5, find a12.

SOLUTION:  

Substitute a1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.

22. If a22 = 95 and a1 = 11, find d.

SOLUTION:  

Substitute a1 = 11, n = 22, and a22 = 95 into the formula for the nth term of an arithmetic sequence.

23. Find a6 for 84, 5, −74, …

SOLUTION:  5 – 84 = –79 –74 – 5 = –79   Substitute a1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.

24. If an = −20, a1 = 46, and d = −11, find n.

SOLUTION:  

Substitute a1 = 46, d = –11, and an = −20 into the formula for the nth term of an arithmetic sequence.

25. If a35 = −63 and a1 = 39, find d.

SOLUTION:  

Substitute a1 = 39, n = 35, and a35 = −63 into the formula for the nth term of an arithmetic sequence.

26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let an represent the number of

pickets in n sections. a. Find the first five terms of the sequence. b. Write a recursive formula for the sequence in part a. c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?  

SOLUTION:  a. If each section contains 14 pickets, a1 = 14 and d = 14.

14 + 14 = 28 28 + 14 = 42 42 + 14 = 56 56 + 14 = 70   Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.   b. Substitute a1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.

  c. Find the number of sections n. Substitute an = 448,

a1 = 14, and d = 14 into the formula you found in part b.

  Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.

Find the indicated arithmetic means for each set of nonconsecutive terms.27. 3 means; 19 and −5

SOLUTION:  

The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a5.

  First, find the common difference using a5 = –5, a1 = 19, and n = 5.

  Next, determine the arithmetic means by using d = –6. 19 + (–6) = 13 13 + (–6) = 7 7 + (–6) = 1   Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.

28. 5 means; –62 and −8

SOLUTION:  The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of

the sequence or a7.

  First, find the common difference using a7 = –8, a1 = –62, and n = 7.

  Next, determine the arithmetic means by using d = 9. –62 + 9 = –53 –53 + 9 = –44 –44 + 9 = –35 –35 + 9 = –26 –26 + 9 = –17   Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.

29. 4 means; 3 and 88

SOLUTION:  The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence

or a6.

  First, find the common difference using a6 = 88, a1 = 3, and n = 6.

  Next, determine the arithmetic means by using d = 17. 3 + 17 = 20 20 + 17 = 37 37 + 17 = 54 54 + 17 = 71   Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.

30. 8 means; −5.5 and 23.75

SOLUTION:  The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note

that 23.75 is the tenth term of the sequence or a10.

  First, find the common difference using a10 = 23.75, a1 = –5.5, and n = 10.

  Next, determine the arithmetic means by using d = 3.25. –5.5 + 3.25 = –2.25 –2.25 + 3.25 = 1 1 + 3.25 = 4.25 4.25 + 3.25 = 7.5 7.5 + 3.25 = 10.75 10.75 + 3.25 = 14 14 + 3.25 = 17.25 17.25 + 3.25 = 20.5   Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14, 17.25, 20.5, 23.75.

31. 7 means; −4.5 and 7.5

SOLUTION:  The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is

the ninth term of the sequence or a9.

  First, find the common difference using a9 = 7.5, a1 = –4.5, and n = 9.

  Next, determine the arithmetic means by using d = 1.5. –4.5 + 1.5 = –3 –3 + 1.5 =  –1.5 –1.5 + 1.5 = 0 0 + 1.5 = 1.5 1.5 + 1.5 = 3 3 + 1.5 = 4.5 4.5 + 1.5 = 6   Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.

32. 10 means; 6 and 259

SOLUTION:  The sequence will resemble  6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__

259. Note that 259 is the eleventh term of the sequence or a12.

  First, find the common difference using a12 = 259, a1 = 6, and n = 12.

  Next, determine the arithmetic means by using d = 23. 6 + 23 or 29 39 23 or 52 52 23 or 75 7 + 23 or 98 98 + 23 or 121 121 + 23 or  144 144 + 23 or 167 167 + 23 or 190 190 + 23 or 213   Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213, 236.

Find a quadratic model for each sequence.33. 12, 19, 28, 39, 52, 67, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 4n + 7c.

34. −11, −9, −5, 1, 9, 19, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH menu to find the solution.

  Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for an, the model for the sequence

is an = n2 – n – 11.

35. 8, 3, −6, −19, −36, −57, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for an, the model for the sequence is

an = –2n2 + n + 9.

36. −7, −2, 9, 26, 49, 78, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for an, the model for the sequence is

an = 3n2 – 4n – 6.

37. 6, −2, −12, −24, −38, −54, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for an, the model for the sequence

is an = –n2 – 5n + 12.

38. −3, 1, 13, 33, 61, 97, …

SOLUTION:  

The nth term can be represented by a quadratic equation of the form an = an2 + bn + c. Substitute values for an

and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the solution.

  Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for an, the model for the sequence is

an = 4n2 – 8n + 1.

Find the indicated sum of each arithmetic series.39. 26th partial sum of 3 + 15 + 27 + … + 303

SOLUTION:  

In this sequence, a1 = 3 and d = 15 – 3 or 12. Find the sum of the series.

40. −28 + (−19) + (−10) + … + 242

SOLUTION:  

In this sequence, a1 = –28, an = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

41. 42nd partial sum of 120 + 114 + 108 + …

SOLUTION:  

In this sequence, a1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.

42. 54th partial sum of 213 + 205 + 197 + …

SOLUTION:  

In this sequence, a1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.

43. −17 + 1 + 19 + … + 649

SOLUTION:  

In this sequence, a1 = –17, an = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

44. 89 + 58 + 27 + … + (–562)

SOLUTION:  

In this sequence, a1 = 89, an = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of

terms in the sequence n.

  Find the sum of the series.

45. RUNNING Refer to the beginning of the lesson. a. Determine the number of miles Meg will run on her 12th day of training. b. During which day of training will Meg reach her goal of 100 total miles?

SOLUTION:  a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,

and so on. Therefore, a1 = 1, a2 = 1.25, and a3 = 1.5. Find the common difference.

1.25 – 1 = 0.25 1.5 – 1.25 = 0.25   The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a12.

 

So, Meg will run 3.75 miles on her 12th

day of training.

  b. Find the term that corresponds to Sn = 100.

  First, write an explicit formula using a1 = 1 and d = 0.25.

  Substitute Sn = 100, a1 = 1, and an = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.

  Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.

Find the indicated sum of each arithmetic series.

46. 

SOLUTION:  

  The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a1 = 5, an = 43, and n = 20. Find the sum of the series.

47. 

SOLUTION:  

  The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a1 = 96, an = –12, and n = 28. Find the sum of the

series.

48. 

SOLUTION:  

  The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a1 = –89, an = –188, and n = 12. Find the

sum of the series.

49. 

SOLUTION:  

  The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a1 = 43, an = 365, and n = 47. Find the sum of the

series.

50. 

SOLUTION:  

  The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 42 – 7 + 1 or 36. Therefore, a1 = 63, an = –42, and n = 36. Find the sum of the

series.

51. 

SOLUTION:  

  The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus

the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a1 = 32, an = 80, and n = 13. Find the sum of the

series.

52. 

SOLUTION:  

The first term of this series is 1 and the last term is 3. The number of terms is equal to the upper bound minus the

lower bound plus one, which is 24 – 20 + 1 or 5. Therefore, a1 = 3, an = 3, and n = 5. Find the sum of the series.

53. 

SOLUTION:  

  The first term of this series is –42 and the last term is –147. The number of terms is equal to the upper bound

minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a1 = –42, an = –147, and n = 8. Find the sum

of the series.

54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number of tiles needed to create the mosaic design.

SOLUTION:  

The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a1 = 8, d = 2, and n =

20. Find the sum of the series.

  Therefore, 540 tiles are needed to create the mosaic design.

55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?  

   

SOLUTION:  The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th

second. So, a1 = 12, d = 1.5, and n = 64. Find the sum of the series.

  Therefore, the competitor travels 3792 feet in 64 seconds.

56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the walk have generated a total of at least $65,000?

SOLUTION:  The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900

each year, so a1 = 3000 and d = 900.

  Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.

57. Find an if Sn = 490, a1 = –5, and n = 100.

SOLUTION:  

Substitute Sn = 490, a1 = –5, and n = 100 into the formula for the sum of an arithmetic series.

58. If Sn = 51.7, n = 22, an = –11.3, find a1.

SOLUTION:  

Substitute Sn = 51.7, n = 22, and an = –11.3 into the formula for the sum of an arithmetic series.

 

59. Find n for –7 + (–5.5) + (–4) + … if Sn = –14 and an = 3.5.

SOLUTION:  

Substitute Sn = –14, an = 3.5, and a1 = –7 into the formula for the sum of an arithmetic series.

 

60. Find a1 if Sn = 1287, n = 22, and d = 5.

SOLUTION:  

Substitute Sn = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.

61. If S26 = 1456, and a1 = –19, find d

SOLUTION:  

Substitute S26 = 1456, n = 26, and a1 = –19 into the formula for the sum of an arithmetic series.

62. If S12 = 174, a12 = 39, find d.

SOLUTION:  

Substitute a12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a1.

  Substitute S12 = 174, a1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic

series.

Write each arithmetic series in sigma notation. The lower bound is given.63. 6 + 12 + 18 + … + 66; n = 1

SOLUTION:  First, find the common difference. 12 – 6 = 6 18 – 12 = 6   Next, find n.

  To find an explicit formula, substitute a1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as  

64. –1 + 0 + 1 + … + 7; n = 1

SOLUTION:  First, find the common difference. 0 – (–1) = 1 1 – 0 = 1   Next, find n.

  To find an explicit formula, substitute a1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series –1 + 0 + 1 + . . . 7 can be described as .

65. 17 + 21 + 25 + … + 61; n = 4

SOLUTION:  First, find the common difference. 21 – 17 = 4 25 – 21 = 4   Find a1.

17 –  4 =13 13 –  4 = 9 9 –  4 = 5 Thus, a1

= 5.

  Next, find n.

  Substitute a1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.

 

Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as .

66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6

SOLUTION:  First, find the common difference. 0 – 1 or –1 –1 – 0 or –1   Find a1.

1 – (–1) = 2 2 – (–1) = 3 3 – (–1) = 4 4 – (–1) = 5 5 – (–1) = 6

Thus, a1 = 6.

  Next, find n. 

  Substitute a1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series 1 + 0 + (–1) + (–2) + . . . + (–13) can be described as  .

67. 

SOLUTION:  First, find the common difference.

  Find a1.

a1 =

  Next, n.

 

Substitute a1 = and d = in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the series can be described as

.

68. 9.25 + 8.5 + 7.75 + … – 2; n = 1

SOLUTION:  First, find the common difference. 8.5 – 9.25 = –0.75 7.75 – 8.5 = –0.75   Next, find n.

  To find an explicit formula, substitute a1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic

sequence.

Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + –2 can be described as   .

69. CONCERTS The seating in a concert auditorium is arranged as shown below.  

  a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown in the first 3 rows continues for each successive row. b. Find the total number of seats in the auditorium. c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows. How many seats are there in this auditorium?

SOLUTION:  a. From the seating chart, a1 = 24, a2 = 29, a3 = 34, and n = 35. Find the common difference.

29 – 24 = 5 34 – 29 = 5   To find an explicit formula, substitute a1 = 24 and d = 5 in the formula for the nth term of an arithmetic sequence.

Therefore, in sigma notation, the number of seats in the auditorium can be described as   .

  b. From part a, a1 = 24 and d = 5. Find S35.

  Therefore, there are 3815 seats in the auditorium.   c. For the second auditorium, n = 32, a1 = 18, and d = 4. Find S32.

  Therefore, there are 2560 seats in the second auditorium.

Write a function that can be used to model the nth term of each sequence.

70. 2, 5, 8, 11, 14, 17, …

SOLUTION:  Find the first differences.

  The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for the nth term of an arithmetic sequence.

71. 8, 13, 20, 29, 40, 53, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = 2, and c = 5. Substituting these values in the equation for an, the model for the sequence is an

= n2 + 2n + 5.

72. 2, 2, 4, 8, 14, 22, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.  

  The second differences are constant, so the sequence can be modeled by a quadratic equation of the form an =

an2 + bn + c. Substitute values for an and n into the equation.

  This yields a system of linear equations in three variables.

  You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –3, and c = 4. Substituting these values in the equation for an, the model for the sequence is

an = n2 – 3n + 4.

73. 5, 31, 97, 221, 421, 715, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for an, the model for the

sequence is an = 3n3 + 2n

2 – n + 1.

74. –6, –8, –6, 6, 34, 84, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

 

The third differences are constant, so the sequence can be modeled by a cubic equation of the form an = an3 +

bn2 + cn + d. Substitute values for an and n into the equation.

  This yields a system of linear equations in four variables.

  You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for an, the model for the

sequence is an = n3 – 4n

2 + 3n – 6.

75. 0, 23, 134, 447, 1124, 2375, …

SOLUTION:  Find the first differences.

  The first differences are not constant. Find the second differences.

  The second differences are not constant. Find the third differences.

  The third differences are not constant. Find the fourth differences.

 

The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form an = an4 +

bn3 + cn

2 + dn + e. Substitute values for an and n into the equation.

  This yields a system of linear equations in five variables.

  You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds to the system.

  Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH menu to find the solution.

  Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for an, the model for the

sequence is an = 2n4 – n

3 – 1.

Find each common difference.

76. 

SOLUTION:  

Find a1 and a100.

 

  14 – 8 = 6   The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.

77. 

SOLUTION:  

Find a21 and a65.

 

  The expression in sigma notation is linear, so the common difference is the coefficient of the variable, .

78. a12 = 63, a19 = 7

SOLUTION:  

a19 is 7 terms from a12.

   

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 46

10-2 Arithmetic Sequences and Series

Page 47: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 47

10-2 Arithmetic Sequences and Series

Page 48: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 48

10-2 Arithmetic Sequences and Series

Page 49: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

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10-2 Arithmetic Sequences and Series

Page 50: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 50

10-2 Arithmetic Sequences and Series

Page 51: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 51

10-2 Arithmetic Sequences and Series

Page 52: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 52

10-2 Arithmetic Sequences and Series

Page 53: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 53

10-2 Arithmetic Sequences and Series

Page 54: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

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Page 55: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

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Page 56: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 56

10-2 Arithmetic Sequences and Series

Page 57: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 57

10-2 Arithmetic Sequences and Series

Page 58: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 58

10-2 Arithmetic Sequences and Series

Page 59: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

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Page 60: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

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Page 61: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 61

10-2 Arithmetic Sequences and Series

Page 62: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

eSolutions Manual - Powered by Cognero Page 62

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Page 63: , the model for the sequence · Determine the common difference, and find the next four terms of each arithmetic sequence. 20, 17, 14, « 62/87,21 First, find the common difference

 

79. a8 = −4, a27 =

SOLUTION:  

a27 is 19 terms from a8.

   

80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using

sequences. Consider f (x) = x2 on the interval [1, 3].

a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.

b. Write the sequence yn formed when yn = f (xn).

c. Write the sequence p n defined by d ⋅ yn.

d. The left-hand approximation of the area is given by . Find L6.

e . The right-hand approximation of the area is given by Find R6.

SOLUTION:  a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,

where 3 is x7.

  First, find the common difference using x7 = 3, x1 = 1, and n = 7.

 

Next, determine the arithmetic means by using d =

1 +  = 

 = 

 = 2

2 +  = 

 = 

 

Therefore, a sequence with three arithmetic means between 1 and 3 is

  b. Find each term in the sequence formed when yn = f (xn).

 

 

 

 

 

 

    Therefore, the sequence formed when yn = f (xn) is

  c. Next, find each term in the sequence defined by d ⋅ yn.

 

 

 

 

 

 

 

  Therefore, the sequence defined by d ⋅ yn is

  d.

  e .

81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit

formula an = 2 + 7(n − 1) for the sequence. Candace’s formula is an = 7n − 5. Is either of them correct? Explain.

SOLUTION:  Write an explicit formula for the sequence. First, find the common difference. 9 – 2 = 7 16 – 9 = 7   Next, substitute a1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.

  Sample answer: Candace's formula an = 7n − 5 is the simplified form of Peter's formula an = 2 + 7(n − 1).

Therefore, both Candace and Peter are correct.

82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes, provide an example. If no, how can the sequence be modeled? Explain.

SOLUTION:  Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function.  Consider the sequence a1, a2, a3, … . If the common difference of the sequence is 0, then the partial sums Sn of

the corresponding series can be modeled by a linear function of the form Sn = a1n.

Otherwise, the partial sums can be modeled by a quadratic function of the form where d

is the common difference of the original sequence.

83. CHALLENGE Prove that for an arithmetic sequence, an = ak + (n – k)d for integers k in the domain of the

sequence.

SOLUTION:  

Sample answer: The formula for the nth term of an arithmetic sequence is an = a1 + (n – 1)d. Similarly, ak = a1 +

(k – 1)d. Solving the second equation for a1 and substituting into the first yields:

REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.

84. If you know the sum and d, you can solve for a1.

SOLUTION:  

The is statement "If you know the sum and d, you can solve for a1." is false. You must also know n.

Consider both formulas for the sum of a finite arithmetic series.  and . In the first formula, you must know Sn, n, and an to find a1. In

the second formula, you must know Sn, n, and an

and d to find a1.

85. If you only know the first and last terms, then you can find the sum.

SOLUTION:  The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know n.   Consider the two formulas for the sum of a finite arithmetic series. 

and . For the first formula, in addition to the first and last terms, you

also need n. In the second formula, you need the first term, n, and d. 

86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of the series is positive.

SOLUTION:  The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positiveor the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3 terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.   Consider the sequence 33, 24, 15, ... Find the common difference. 24 – 33 = –9 15 – 24 = –9   Use the common difference of –9 to find the next three terms. 15 – 9 = 6 6 – 9 = –3 –3  – 9 = –12 The terms in the sequence are no all positive.    Find the sum of the first 10 terms in the sequence. 

The sum is negative. 

87. CHALLENGE Consider the arithmetic sequence of odd natural numbers. a. Find S7 and S9.

b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series. c. Write an algebraic proof verifying the conjecture that you made in part b.

SOLUTION:  a. The arithmetic sequence of odd natural numbers is a1 = 1, a2 = 3, a3 = 5, a5 = 7, a6 = 9, a7 = 11, a8 = 13, a9 =

15… . The common difference is 2.   Find S7.

  Find S9.

  b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum

of the first n terms of the sequence of odd natural numbers appears to be n2.

  c. Write an explicit formula for the arithmetic sequence of odd natural numbers.

  Substitute an = 2n – 1 and a1 = 1 into the formula for the sum of a finite arithmetic series.

88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.

SOLUTION:  Sample answer:

 

The partial sum Sn is  which approaches – as n   . Therefore, the sum of the series 25 + 20

+ 15 + … cannot be calculated.

Find the next four terms of each sequence.89. 12, 16, 20, …

SOLUTION:  These terms appear to increase by 4. Check.16 – 12 =  4 20 – 16 = 4   The next four terms are  20 + 4 =  24 24 + 4 =  28 28 + 4 =  32 32 + 4 =  36.

90. 3, 1, −1, …

SOLUTION:  The terms appear to decrease by 2. 1 – 3 = –2 –1 – 1 = –2   The next four terms are : –1 –2 = –3 –3 –2 = –5 –5 –2 = –7 –7 –2 = –9.

91. 31, 24, 17, …

SOLUTION:  These terms appear to decrease by 7. Check.  24 – 31 = –7 17 – 24 = –7   The next four terms are:   17 – 7 = 10 10 – 7 = 3 3 – 7 = –4 –4 – 7 = –11 .

Find each product or quotient and express it in rectangular form.

92. 

SOLUTION:  Use the Product Formula to find the product in polar form.

Now find the rectangular form of the product.

The polar form of the product is . The rectangular form of the quotient is .

93. 

SOLUTION:  Use the Quotient Formula to find the quotient in polar form.

Now find the rectangular form of the product.

The polar form of the quotient is . The rectangular form of the quotient is .

Find the dot product of u and v. Then determine if u and v are orthogonal.94. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

95. u = , v =  

SOLUTION:  

Since , u and v are not orthogonal.

96. u = , v =

SOLUTION:  

Since , u and v are not orthogonal.

Find the direction angle of each vector to the nearest tenth of a degree.97. –i – 3j

SOLUTION:  

Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.

98. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.

99. 

SOLUTION:  

Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.

100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation  +   = 1. The 

camshaft goes through the focus on the positive axis. a. Graph a model of the cam. b. Find an equation that translates the model so that the camshaft is at the origin. c. Find the equation of the model in part b when the cam is rotated to an upright position.

SOLUTION:  a.

  b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.

 

The right focus is located at ( , 0), so the graph of   needs to be shifted  units to the left for 

the focus to be at the origin. Therefore, the translated equation is .

  c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.   First, find the equations for x and y .

 

  Substitute these values into the original equation.

  Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by

.

101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then sketch the graphs of f and g.

SOLUTION:  This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the right and expanded vertically.  

102. SAT/ACT What is the units digit of 336

?

  A 0 B 1 C 3 D 7 E 9

SOLUTION:  

Find 3n for whole number values of n, where 1 ≤ n ≤ 8.

 

 

Notice that there is a pattern emerging: 31 = 3 and 3

5 = 243, 3

2 = 9 and 3

6 = 729, 3

3 = 27 and 3

7 = 2187, and 3

4 =

81 and 38 = 6561. From this pattern, you can conclude that 3

36 will also have a units digit of 1. Therefore, the

correct answer is B.

103. Using the table, which formula can be used to determine the nth term of the sequence?

  F an = 6n

G an = n + 5

H an = 2n + 1

J an = 4n + 2

SOLUTION:  

From the table, a1 = 6, a2 = 10, a3 = 14, and a4 = 18. The common difference is 10 – 6 or 4.

  Write an explicit formula for the nth term of the sequence.

  Therefore, the correct answer is J.

104. REVIEW If a1 = 3, a2 = 5, and an = an − 2 + 3n, find a10.

  A 59 B 75 C 89 D 125

SOLUTION:  

Substitute n = 10 into an = an − 2 + 3n.

  To find a10, you must find a8. However, to find a8, you must first find a3, a4, …, a6.

 

 

 

  Now that a6 is known, you can find a8.

  Substitute a8 = 59 into the equation for a10.

  Therefore, the correct answer is C.

105. REVIEW Which of the sequences shown below is convergent?  

SOLUTION:  A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that onlythe sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct answer is J.

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10-2 Arithmetic Sequences and Series