Download ppt - Course 3 13-1 Terms of Arithmetic Sequences 13-1 Terms of Arithmetic Sequences Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation

Course 3

13-1 Terms of Arithmetic Sequences13-1 Terms of Arithmetic Sequences

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Course 3

13-1 Terms of Arithmetic Sequences

Warm UpFind the next two numbers in the pattern, using the simplest rule you can find.

1. 1, 5, 9, 13, . . .

2. 100, 50, 25, 12.5, . . .

3. 80, 87, 94, 101, . . .

4. 3, 9, 7, 13, 11, . . .

17, 21

6.25, 3.125

108, 115

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17, 15

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Problem of the Day

Write the last part of this set of equations so that its graph is the letter W.y = –2x + 4 for 0 x 2y = 2x – 4 for 2 < x 4y = –2x + 12 for 4 < x 6

Possible answer: y = 2x – 12 for 6 < x 8

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Learn to find terms in an arithmetic sequence.

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You cannot tell if a sequence is arithmetic by looking at a finite number of terms because the next term might not fit the pattern.

Caution!

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Determine if the sequence could be arithmetic. If so, give the common difference.

5, 8, 11, 14, 17, . . .

Additional Example 1A: Identifying Arithmetic Sequences

The terms increase by 3.

The sequence could be arithmetic with a common difference of 3.

5 8 11 14 17, . . .

3333

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1, 3, 6, 10, 15, . . .

Additional Example 1B: Identifying Arithmetic Sequences

The sequence is not arithmetic.

Find the difference of each term and the term before it.

1 3 6 10 15, . . .

5432

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65, 60, 55, 50, 45, . . .

Additional Example 1C: Identifying Arithmetic Sequences

The sequence could be arithmetic with a common difference of –5.

The terms decrease by 5.65 60 55 50 45, . . .

–5–5–5–5

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5.7, 5.8, 5.9, 6, 6.1, . . .

Additional Example 1D: Identifying Arithmetic Sequences

The sequence could be arithmetic with a common difference of 0.1.

The terms increase by 0.1.5.7 5.8 5.9 6 6.1, . . .

0.10.10.10.1

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1, 0, -1, 0, 1, . . .

Additional Example 1E: Identifying Arithmetic Sequences



1 0 –1 0 1, . . .

11–1–1

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1, 2, 3, 4, 5, . . .

Check It Out: Example 1A


The terms increase by 1.1 2 3 4 5, . . .

1111

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1, 3, 7, 8, 12, …

Check It Out: Example 1B



1 3 7 8 12, . . .

4142

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11, 22, 33, 44, 55, . . .

Check It Out: Example 1C


The terms increase by 11.11 22 33 44 55, . . .

11111111

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1, 1, 1, 1, 1, 1, . . .

Check It Out: Example 1D



1 1 1 1 1, . . .

0000

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2, 4, 6, 8, 9, . . .

Check It Out: Example 1E



2 4 6 8 9, . . .

1222

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Helpful Hint

Subscripts are used to show the positions of terms in the sequence. The first term is a1, “read a sub one,” the second is a2, and so on.

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Find the given term in the arithmetic sequence.

10th term: 1, 3, 5, 7, . . .

Additional Example 2A: Finding a Given Term of an Arithmetic Sequence

an = a1 + (n – 1)d

a10 = 1 + (10 – 1)2

a10 = 19

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18th term: 100, 93, 86, 79, . . .

Additional Example 2B: Finding a Given Term of an Arithmetic Sequence

an = a1 + (n – 1)d

a18 = 100 + (18 – 1)(–7)

a18 = -19

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21st term: 25, 25.5, 26, 26.5, . . .

Additional Example 2C: Finding a Given Term of an Arithmetic Sequence

an = a1 + (n – 1)d

a21 = 25 + (21 – 1)(0.5)

a21 = 35

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14th term: a1 = 13, d = 5

Additional Example 2D: Finding a Given Term of an Arithmetic Sequence

an = a1 + (n – 1)d

a14 = 13 + (14 – 1)5

a14 = 78

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15th term: 1, 3, 5, 7, . . .

Check it Out: Example 2A

an = a1 + (n – 1)d

a15 = 1 + (15 – 1)2

a15 = 29

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50th term: 100, 93, 86, 79, . . .

Check It Out: Example 2B

an = a1 + (n – 1)d

a50 = 100 + (50 – 1)(-7)

a50 = –243

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41st term: 25, 25.5, 26, 26.5, . . .

Check It Out: Example 2C

an = a1 + (n – 1)d

a41 = 25 + (41 – 1)(0.5)

a41 = 45

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2nd term: a1 = 13, d = 5

Check It Out: Example 2D

an = a1 + (n – 1)d

a2 = 13 + (2 – 1)5

a2 = 18

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You can use the formula for the nth term of an arithmetic sequence to solve for other variables.

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The senior class held a bake sale. At the beginning of the sale, there was $20 in the cash box. Each item in the sale cost 50 cents. At the end of the sale, there was $63.50 in the cash box. How many items were sold during the bake sale?

Additional Example 3: Application

Identify the arithmetic sequence: 20.5, 21, 21.5, 22, . . .

a1 = 20.5 a1 = 20.5 = money after first sale

d = 0.5

an = 63.5

d = .50 = common difference

an = 63.5 = money at the end of the sale

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Additional Example 3 ContinuedLet n represent the item number of cookies sold that will earn the class a total of $63.50. Use the formula for arithmetic sequences.

an = a1 + (n – 1) d

Solve for n.63.5 = 20.5 + (n – 1)(0.5)

63.5 = 20.5 + 0.5n – 0.5 Distributive Property.

63.5 = 20 + 0.5n Combine like terms.

87 = n

Subtract 20 from both sides.

During the bake sale, 87 items are sold in order for the cash box to contain $63.50.

43.5 = 0.5n

Divide both sides by 0.5.

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Johnnie is selling pencils for student council. At the beginning of the day, there was $10 in his money bag. Each pencil costs 25 cents. At the end of the day, he had $40 in his money bag. How many pencils were sold during the day?

Check It Out: Example 3

Identify the arithmetic sequence: 10.25, 10.5, 10.75, 11, …

a1 = 10.25

d = 0.25

an = 40

a1 = 10.25 = money after first sale

d = .25 = common difference

an = 40 = money at the end of the sale

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Check It Out: Example 3 ContinuedLet n represent the number of pencils in which he will have $40 in his money bag. Use the formula for arithmetic sequences.an = a1 + (n – 1)d

Solve for n.40 = 10.25 + (n – 1)(0.25)

40 = 10.25 + 0.25n – 0.25 Distributive Property.

40 = 10 + 0.25n Combine like terms.

120 = n

Subtract 10 from both sides.

120 pencils are sold in order for his money bag to contain $40.

30 = 0.25n

Divide both sides by 0.25.

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Lesson QuizDetermine if each sequence could be arithmetic. If so, give the common difference.

1. 42, 49, 56, 63, 70, . . .

2. 1, 2, 4, 8, 16, 32, . . .

Find the given term in each arithmetic

sequence.

3. 15th term: a1 = 7, d = 5

4. 24th term: 1, , , , 2

5. 52nd term: a1 = 14.2; d = –1.2

no

yes; 7

77

54

32

74

, or 6.7527 4

–47