1.1

Warm-up Practice:

A.

1. 1, 3, 5, 7; 19; 29 2. 5, 8, 11, 14; 32; 47

3. 3, 6, 11, 18; 102; 227 4. -2, 1, 6, 13; 97; 222

5. , 2/3, , 4/5; 10/11; 15/16 6. 1/5, 1/3, 3/7, 1/2; 5/7; 15/19

7. 2, 3/2, 4/3, 5/4; 11/10; 16/15 8. 0, , 2/3, ; 9/10; 14/15

9. 0, , 2/5, ; ; 14/17 10. , 4/5, 5/6, 6/7; 12/13; 17/18

11. -1, 2, -3, 4; 10; -15 12. -1, 2, -3, 4; 10; -15

B.

13. 32 14. 33

15. 208 16. 9/10

C.

17. An = 2n 1 18. An = 2n

19. An =

; An = 1 +

20. An =

An =

21. An = 3n 2 22. An =

; An = 1

23. An = 2n x (-1)n - 1

;An = 2n x (-1)n + 1

24. An = 3n x (-1)^(n 1);An = 3n x (-1)n + 1

25. An = -3n + 2 26. An = -4n + 3

27. An = (5)n; 5

n/2 28. An = n

2 + n

D.

29. 63 30. 126

31. 43/10 or 4 3/10 32. 17/4, 4 1/4

33. -13/15 34. 5/12

E.

35. 2 + 4 + 6 + 8 + 10 + 12 = 42 36. 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 = 92

37. 0 + + 2/3 + + 4/5 = 163/60; 2 43/60 38. 1 + 1 + 1 + 1 +1 = 5

39. -1 + 1 1 + 1 1 + 1 = 0 40. + 1/3 + + 1/5 = 77/60;1 17/60

F.

41. 2x, 4x, 6x, 8x, 10x 42. x, x2, x

3, x

4, x

5, x

6

43. 1/x3, 1/x

4, 1/x

5, 1/x

6, 1/x

7, 1/x

8 44. x, 2x

2, 3x

3, 4x

4, 5x

5

45. 1/x, 4/x2, 9/x

3, 16/x

4, 25/x

5 46.

,

,

,

,

Powerplus

A.

1. (x + 1) + (x + 2) + (x + 3) + (x + 4) 2. (x 3) + (x 4) + (x 5) + (x 6)

3. (x + 1) + (x + 1) + (x + 1)4 + (x + 1)

5 + (x + 1)

6 + (x + 1)

7 4. (x + 3) + (x + 3)

2 + (x + 3)

3 + (x + 3)

4

5.

+

+

+

+

6.

+

+

+

+

+

B.

1. 2. 3.

4.

5.

6.

7.

8.

9.

10.

11. 12. 13.

14.

C.

2 is True. (walang page 14, na ulit yung 13)

D.

a.

n F1 + F2 + . Fn F12 + F2

2 + . Fn

2 Fn + 2 1 FnFn + 1

1 1 1 1 1

2 2 2 2 2

3 4 6 4 6

4 7 15 7 15

5 12 40 12 40

Notice that:

1. F1 + F2 + . Fn = Fn + 2 1

2. F12 + F2

2 + . Fn

2 = FnFn + 1

b.

n F2

n + 1

FnFn + 2 + (-1)n

1 1 1

2 4 4

3 9 9

4 25 25

5 64 64

Yes.

Proof by Induction:

First, for n = 1, . F2

2 = F1F3 + (-1)n

=> 12 = 1(2) 1. This is true.

For n = 2, 22

= 1(3) + 1. Also True.

Next, assume that 1 F2

n + 1 = FnFn + 2 1 is true for some n that is a positive odd integer.

We know that Fn = Fn + 2 Fn + 1 . Subsitute, we get

F2

n + 1 = ( Fn + 2 Fn + 1) Fn + 2 1

F2

n + 1 = F2

n + 2 - Fn + 1Fn + 2 1

F2

n + 2 = F2

n + 1 + Fn + 1Fn + 2 + 1

F2

n + 2 = Fn + 1(Fn + 1 + Fn + 2) + 1

But, Fn + 1 + Fn + 2 = Fn + 3 , so,

2F2

n + 2 = Fn + 1Fn + 3 + 1. So, 1 implies 2. Let n + 1 = a, where a is even, since n is odd. Similarly,

F2

a + 2 = ( Fa + 3 Fa + 2) Fa + 3 + 1

F2

a + 2 = F2

a + 3 - Fa + 2Fa + 3 + 1

F2

a + 2 = F2

a + 3 + Fa + 2Fa + 3 + 1

F2

a + 3 = Fa + 2(Fa + 2 + Fa + 3) 1, but Fa + 2 + Fa + 3 = Fn + 4 .

3. F2

a + 2 = FaFa + 3 2 1. => F2

n + 3 = Fn + 2Fn + 4 1. We know that n + 3 is even. Therefore, the equation will have + 1 when n is

even and 1 when n is odd. We can denote that as:

F2

n + 1 = FnFn + 2 + ( 1)n , where n is a positive integer.

Take the Challenge:

1. 5. (37 + 38, 24 + 25 + 26, 13 + 14 + 15 + 16 + 17, 10 + 11 + 12 + 13 + 14 + 15,3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)

2. 7747

3. 1

4. 5

5. 10

1.2

A.

1. Arithmetic; d = 1 2. Arithmetic; d = 3

3. Not Arithmetic 4. Arithmetic; d = 4

5. Arithmetic; d = 5 6. Not Arithmetic

7. Not Arithmetic 8. Arithmetic; d =

9. Arithmetic; d = 2/3 10. Not Arithmetic

B.

11. an = 5n 3 12. an = 3n + 8

13. an =

14. an = 10n 6

15. an = 3n 6 16. an = 4n 9.4721

C.

17. 76 18. 9/2

19. 16.207 20. 43/5

21. 1 22. 37

23. 5 24. 8.4

D.

25. 16 26. 15

27. 46 28. 13

29. 21 30. 14

E.

31. 10 32. 21

33. 96 34. 44

35. 1220 36. 210

37. 3775 38. 765

F.

42, 44, 46 are true.

Powerplus

A.

1. 8, 13, 18; 88; Sum = 816 2. 2, 8, 14; 116; 1180

3. 2, -1, -4; -85; Sum = -1245 4. 4, 2, 0; -74; -1400

5. 4, 8, 12; 400; Sum = 20,200 6. -2, -4, -6; -200; Sum = -10100

B.

1. 396 2. 504

3. 1275 4. 3488

5. -27 6. -65

7. 3888 8. 320

9. 7 10. 630 cm

11. No. The last row will contain 3 blocks, and there will be a total of 9 rows.

12. 6 13. P7100 14. P465

15.

a.

b.

y(x + z) = 2xz y(x + z) = 2xz

xy + yz = 2xz

xy xz = xz yz

x(y z) = z(x y)

16. Kulang sa info.

17. 579 18. 99

19.

20. Sn =

=

= 3n

2 n

Cancel n on both sides, we get

= 3n 1

2 + an = 6n 2

an = 6n 4

Take the Challenge

1. 40

2. 78

3. 92

4. 166 ( not sure )

5. 45

SECTION 1.3

Warmup Practice

A.

1. Geometric; ratio = 5 2. Geometric; ratio = 2

3. Geometric; ratio = 1/3 4. Not geometric

5. Not geometric 6. Geometric; ratio = -1/3

7. Geometric; ratio = -2 8. Not geometric

9. Geometric; ratio = 1/b 10. Geometric; ratio = x/2

B.

11. 10, 5, 5/2, 5/4, 5/8 12. 12, 4, 4/3, 4/9, 4/27

13. -1/4, , -1, 2, -4 14. -1/16, , -1, 4, -16

15. 3, -9, 27, -81, 243 16. 5, -25, 125, -625, 3125

17.

, 2a, 4b,

18. c, b,

,

,

C.

19. 3, 6, 12, 24, 48 20. -2, 4, -8, 16, -32

21. -5, -1, -1/25, -1/125,-1/625 22. 8, 4, 2, 1,

23. 2, 8, 32, 128, 512 24. 3, 6, 12, 24, 48

25. -3, -12, -36, -192, -768 26. 2, -10, 50, -250, 1250

27. -64, 32, -16, 8, -4 28. -64, -32, -16, -8, -4

29. -64, -32, -16, -8, -4 30. -81, 27, -9, 3, -1

31. 2, 10, 50, 250, 1250 32. -3, 9, -27, 81, -243

D.

33. 728 34. 381

35. 33 36. -728

37. -64.5 38. -85/24 or -3 13/24

E.

39. 1093 40. 55987

41. 547 42. 171

43. 31 44.33/32 or 1 31/32

45. 26/81 46. 2343/1024 or 2 295/1024

Powerplus

A.

1. 1/16 2. -1/125

3. -1 4. -1

5. 32; 62 + 6. 81; 121 +

7. 2 8.

9. 3584 10. 1/32

11. 1/27 12.

13. -3 14.

15. 6, 18, 54; -6, 18, -54 16. 6, 12, 24, 48

17. -16 18. 27

19. Impossible 20. 5

B.

1.

sq. units 2. 2047

3. 163.604 or 40951/250 4. Approximately 1286.831681 m

Take the Challenge

1. 28.

2. First, we know that p =

=

=

. Also, the sum of the integers from 1 to p is equal to

=

=

. By the difference of squares property, we have:

=

. Next, we

have 1 + 9 + 92 + . + 9

n =

=

=

. But, 9

n + 1 = 3

2n + 2 .So, we have:

=

3. -516096

4. a. First, we know that r% = 0.01r. Also,let T0 be the starting value of the quantity. So, r% of T0 = 0.01r x T0. So, T1 = T0 +

0.01r x T0 = T0(1 + 0.01r). Similarly, we have T2 = T1(1 + 0.01r) = T0(1 + 0.01r)2. So, this means that T1 , T2 , T3 , . Tn

form a geometric sequence with common ratio (1 + 0.01r) and Tn = T0(1 + 0.01r)n .

b. 8,144,473.134.

5. 1139

6. No integral value of x.

Section 1.4

Warmup Practice

A.

1. 24 2. 4

3. No sum 4. 3

5. 1/4 6. No sum

7. 8/49 8. 8/7

9. 2 10. -1/7

B.

11. 1/3 12.2/3

13. 31/99 14. 5/11

15. 2 410/999 16.3 1/37

17. 29/225 18. 5/6

C.

19. 20 cm 20.120 cm

21. 30 cm 22. 24 cm

Powerplus

A.

1. -1/2 2.

3. a1 = 1; r = 1/3 4. 21/8

B.

1. d3 = 7.2; d5 = 6.48; d2n 1 = 8 x 0.9(n 1)

2. d4 = 8.1; d2n = 10 x 0.9(n 1)

3. a. 80(1 - 0.9n) b. 100(1 0.9

n)

4. 190

C.

1. a. 1/2n

b. 10 cuts

2. a. P1P2 = a cm;P2P3 =

cm;P3P4 = a cm

b. Since QP1R is an isosceles right triangle, and all the cuts are perpendiculars, then P1P2 = 2P3P4 = 4P5P6 = . So, this

geometric sequence has common ratio of . Also, P2P3 = 2P4P5 = 4P6P7 = . Finally, we have P1P2 = P2P3 . So, we have P1P2 =

P2P3 = 2 P3P4 = 2 P4P5 = . So, this is a geometric sequence with common ratio 1/ or

c.a(2 + ) cm

Take the Challenge

1. a.

cm

2

b.2 +

cm

c. s = or

2. a. OP2 = 5 cm; A1 = 25 cm2

b. A2 =

cm

2

c. i. ratio =

ii. 50 cm2

iii. We know that since OPnPn + 1 is an isosceles triangle, we know that OPn + 1 = PnPn + 1 =

OPn. Also,

A1 = (OPn + 1) (PnPn + 1) = (

OPn

)

2 = ()(OPn

2) = (ln)

2

Also, A1 + A2 + = 4(l12 + l2

2 + ...) = 50

l12 + l2

2 + ...= 50/4 = 25/2

3. 15/8 mm

4. a. 40 cm

b. 40 cm

c. 400/3 cm2

Section 1.5

A.

1. 5/23, 5/26, 5/29 2. 1/11, 1/13, 1/15

3. 1/22, 1/27, 1/32 4. 1/27, 1/31, 1/35

5. 3/7, 3/8, 1/3 6. 4/23, 1/7, 4/33

B.

1. 1/ 79 2. 2/23

3. 1/18 4. Not a harmonic sequence

Powerplus

A.

1. Impossible. The terms of an arithmetic sequence are 2/5, 1/5, 0, - 1/5, - 2/5. However, taking the reciprocals of each will

result in one having an undefined value.

2. 12/11, 12/7, 4, - 12

3. 12/25, 6/19

4. 1/5, 1/8, 1/11

5. 1/10; 2/n

6. , 1/3, , 1/5, .

7. 9 and 1

8. 2

9. 6 and 24

10. It is given

Take the Challenge

1. First, since m and n are in a harmonic progression, then 1/m and 1/n are in an arithmetic progression. Let a be the first term

in that sequence. So,

1/m = a + (n 1)d => 1 = am + dm(n 1), and

1/n = a + ( m 1)d => 1 = an + dn(m 1). Subtracting the second from the first, we get

1 1 = am an + dm(n 1) dn(m 1) => 0 = a(m n) + dmn dm dmn + dn => 0 = a(m n) d(m n) => 0 = (a d)(m n).

We know that m n, so, we have a d = 0 or a = d. Substituting to any of the equations (both will yield the same result, but for

the sake of this solution lets use the 1st

equation.)

1/m = a + (n 1)d = a + (n 1)a = an => a = 1/mn.

Now, using the formula to get the (m + n)th term, we have a + (m + n 1)d = a + (m + n 1)a = a(m + n) =

So, this is the

(m + n)th term of the arithmetic progression so that its reciprocal is the harmonic progression given. Therefore, the (m + n)th

term of the harmonic progression is

.

2. Since a, b and c are terms of a harmonic progression, 1/a, 1/b and 1/c are corresponding terms of an arithmetic progression.

Let such an arithmetic progression have a general formula of a + (k 1)d. So,

1/a = a + (p 1)d.(i)

1/b = a + (q 1)d (ii)

1/c = a + (r 1)d..(iii)

(i) (ii) gives 1/a 1/b = a + pd d a qd + d = (p q)d => (b a) = (p q)d x ab(iv)

(ii) (iii) gives 1/b 1/c = a + qd d a rd + d = (q r)d => (c b) = (q r)d x bc(v)

(iii) (i) gives 1/c 1/a = a + rd d a pr + d = (r p)d => (a c) = (r p)d x ac.(vi)

Adding (iv), (v), and (vi) yields

(b a) + (c b) + (a c) = (p q)d x ab + (q r)d x bc + (r p)d x ac

0 = d[(p q)ab + (q r)bc + (r p)ac]. But d 0, so

bc(q r) + ca(r p) + ab(p q) = 0, which gives out the desired result.

3. 6 and 3

4.

5.

Section 1.6

A.

1. 6 2. 120 3. 2 4. 1

5. 45 6. 84 7. 7 8. 28

9. 20 10. 11 11. 1 12. 105

B.

13. c5 + 10c

4 + 40c

3 + 40c

2 + 80c + 32 14. 81+108a +54a

2 +12a

3 +a

4

15. x6 + 6x

5y + 15x

4y

2 + 20x

3y

3 + 15x

2y

4 + 6xy

5 + y

6 16. x

6 + 12x

5y + 60x

4y

2 + 160x

3y

3 + 240x

2y

4 + 192xy

5 + 64y

6

17. x5

+ 15x4y + 90x

3y

2 +270x

2y

3 + 405xy

4 +243y

5 18.

19. y16

- 4y12

+ 6y8 4y

4 + 1 20. 32x

4 + 96x

3y + 216x

2y

2 + 216xy

3 + 81y

4

21. x8 + 8x

6y + 24x

4y

2 + 32x

2y

3 + 16y

4 22. 8x

6 12x

4y

2 + 6x

2y

4 y

6

23. 27x6 + 54x

4y

4 + 36x

2y

8 + 8y

12 24.

25.

26. x

6 - 6 x

5 + 30x

4 - 40 x

3 + 60x

2 - 24 x + 8

27. y6 - 6 y

5 + 45y

4 - 60 y

3 + 135y

2 - 54 y + 27 28. 8a

3/2 + a

2 + 24a + 32a

1/2 + 16

29. a4/3

+ 24a2/3

+ 8a + 32a1/3

+ 16 30.

C.

31. x16

+ 8x15

+ 28x14

32. a28

+ 14a27

+ 91x26

33. 256a8 + 1024a

7b + 1792a

6b

2 34. a

12 + 36a

11 + 594a

10

35. a8 16a

7b + 112a

6b

2 36. 128a

7 2240a

6 + 16800a

5 37. a

20 + 10a

18b

2 + 45a

16b

4 38. a

28 14a

26b

2 + 91a

24b

4

39. a42

+ 42a41

b + 861a

40b

2 40. a

93 93a

92b + 4278a

91b

2 41. y

7 + 7y

5 + 21y

3 42. y

8 +