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Copyright © Big Ideas Learning, LLC Algebra 2 397All rights reserved. Worked-Out Solutions

Chapter 8

Chapter 8

Chapter 8 Maintaining Mathematical Profi ciency (p. 407)

1. x 3 − 2x y

1 3 − 21 1

2 3 − 22 −1

3 3 − 23 −5

2. x 5x 2 + 1 y

2 5(2)2 + 1 21

3 5(3)2 + 1 46

4 5(4)2 + 1 81

3. x −4x + 24 y

5 −4(5) + 24 4

10 −4(10) + 24 −16

15 −4(15) + 24 −36

4. 7x + 3 = 31 Check 7(4) + 3 =?

31

7x = 28 28 + 3 =?

31

x = 4 31 = 31 ✓

5. 1 — 16

= 4 ( 1 — 2 ) x Check 1 —

16 =

? 4 ( 1 —

2 ) 6

1 — 64

= ( 1 — 2 ) x 1 —

16 =

? 4 ( 1 —

64 )

log1/2 ( 1 — 64

) = log1/2 [ ( 1 — 2 ) x ] 1 —

16 =

1 —

16 ✓

6 = x

6. 216 = 3(x + 6) Check 216 =?

3(66 + 6)

72 = x + 6 216 =?

3(72)

66 = x 216 = 216 ✓

7. 2x + 16 = 144 Check 27 + 16 =?

144

2x = 128 128 + 16 =?

144

log2 (2x) = log2 (128) 144 = 144 ✓

x = 7

8. 1 — 4 x − 8 = 17 Check

1 —

4 (100) − 8 =

? 17

1 — 4 x = 25 25 − 8 =

? 17

x = 100 17 = 17 ✓

9. 8 ( 3 — 4 ) x =

27 —

8 Check 8 ( 3 —

4 ) 3 =?

27 —

8

( 3 — 4 ) x =

27 —

64 8 ( 27

— 64

) =? 27

— 8

log3/4 ( 3 — 4 ) x = log3/4 ( 27

— 64

) 27

— 8 =

27 —

8 ✓

x = 3

10. Sample answer: The difference between the graph of f and

the scatter plot is that f is decreasing whereas the scatter plot

points are increasing. The similarity between the two is that

both f and the scatter plot level off as x increases.

Chapter 8 Mathematical Practices (p. 408)

1. ATime

(hours)

00.5

11.5

22.5

33.5

4

BDistance(miles)

0250500750

10001250150017502000

1

23456789

1011

From the spreadsheet, it follows that the pattern is linear.

2. AYear

012345678

BPopulation

607594

117146183229286358

123456789

1011

From the spreadsheet, it follows that the pattern is

exponential and it represents exponential growth.

3. ADecade

012345678

BPopulation

500450405365328295266239215

123456789

1011

From the spreadsheet, it follows that the decline is

exponential.

398 Algebra 2 Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 8

4. APlace

12345678

BPrize

$200$175$150$125$100$75$50$25

12345678910

From the spreadsheet, it follows that the decline is linear.

8.1 Explorations (p. 409)

1. a. The graph is B.

The nth term of the sequence is an = 1.5n − 0.5.

When n = 10, a10 = 1.5(10) − 0.5 = 14.5.

b. The graph is D.

The nth term of the sequence is an = −1.5n + 9.5.

When n = 10, a10 = −1.5(10) + 9.5 = −5.5.

c. The graph is A.

The nth term of the sequence is an = 1 —

4 n2.

When n = 10, a10 = 1 —

4 (10)2 = 25.

d. The graph is F.

The nth term of the sequence is an = 1 —

4 (−n + 6)2.

When n = 10, a10 = 1 —

4 (−10 + 6)2 = 4.

e. The graph is C.

The nth term of the sequence is an = 2n − 2.

When n = 10, a10 = 210 − 2 = 256.

f. The graph is E.

The nth term of the sequence is an = 2−n + 4.

When n = 10, a10 = 2−10 + 4 = 1 —

64 .

2. Sample answer: Graph the terms of the sequence, look at the

pattern, and write a rule for the function that is represented

by the graph.

3. a. In an arithmetic sequence the difference between two

consecutive terms is constant.

b. In a geometric sequence the ratio of two consecutive

terms is constant.

8.1 Monitoring Progress (pp. 410–413)

1. a1 = 1 + 4 = 5 2. f (1) = (−2)1 − 1 = 1

a2 = 2 + 4 = 6 f (2) = (−2)2 − 1 = −2

a3 = 3 + 4 = 7 f (3) = (−2)3 − 1 = 4

a4 = 4 + 4 = 8 f (4) = (−2)4 − 1 = −8

a5 = 5 + 4 = 9 f (5) = (−2)5 − 1 = 16

a6 = 6 + 4 = 10 f (6) = (−2)6 − 1 = −32

3. a1 = 1 —

1 + 1 = 1 —

2

a2 = 2 —

2 + 1 =

2 —

3

a3 = 3 —

3 + 1 =

3 —

4

a4 = 4 —

4 + 1 =

4 —

5

a5 = 5 —

5 + 1 = 5 —

6

a6 = 6 —

6 + 1 =

6 —

7

4. You can write the terms as: 3 = 2(1) + 1,

5 = 2(2) + 1,

7 = 2(3) + 1,

9 = 2(4) + 1, . . . .

The fi fth term is a5 = 2(5) + 1 = 11.

A rule for the nth term is an = 2n + 1.

n

an

8

10

12

14

4

6

2

04 5 63210

5. You can write the terms as: 3 = 1(1 + 2),

8 = 2(2 + 2),

15 = 3(3 + 2),

24 = 4(4 + 2), . . . .

The fi fth term is a5 = 5(5 + 2) = 35.

A rule for the nth term is an = n(n + 2).

n

an

24

30

36

12

18

6

04 5 63210

6. You can write the terms as: 1 = (−2)1 − 1,

−2 = (−2)2 − 1,

4 = (−2)3 − 1,

−8 = (−2)4 − 1, . . . .

The fi fth term is a5 = (−2)5 − 1 = 16.

A rule for the nth term is an = (−2)n − 1.

n

an

16

8

12

4

−8

−4

−12

4 5 6 7 831


Chapter 8

7. You can write the terms as: 2 = 12 + 1,

5 = 22 + 1,

10 = 32 + 1,

17 = 42 + 1, . . . .

The fi fth term is a5 = 52 + 1 = 26.

A rule for the nth term is an = n2 + 1.

n

an

24

30

36

12

18

6

04 5 63210

8. Because the nth term of the sequence is an = n2, when

n = 9, a 9 = 92 = 81.

So, there are 81 apples in the ninth layer.

9. Notice that the terms of the series have the pattern

5(1), 5(2), 5(3), . . ., 5(20).

So, the terms of the series can be written as:

ai = 5i, where i = 1, 2, 3, . . ., 20

The lower limit of summation is 1 and the upper limit is 20.

So, the summation notation for the series is ∑ i = 1

20

5i .


12 —

12 + 1 ,

22

— 22 + 1

, 32

— 32 + 1

, 42

— 42 + 1

, . . . .


ai = i2 —

i 2 + 1 , where i = 1, 2, 3, 4, . . . .

The lower limit of summation is 1 and the upper limit is

infi nity.


∞

i 2 —

i 2 + 1 .


61, 62, 63, 64, . . . .


ai = 6i, where i = 1, 2, 3, 4, . . . .


infi nity.


∞ 6i .


1 + 4, 2 + 4, 3 + 4, . . ., 8 + 4


ai = i + 4, where i = 1, 2, 3, . . ., 8.



8

(i + 4) .

13. ∑ i = 1

5

8i = 8(1) + 8(2) + 8(3) + 8(4) + 8(5)

= 8 + 16 + 24 + 32 + 40

= 120

14. ∑ k = 3

7

(k 2 − 1) = (32 − 1) + (42 − 1) + (52 − 1) +

(62 − 1)+ (72 − 1)

= 8 + 15 + 24 + 35 + 48

= 130

15. ∑ i = 1

34

1 = 34

16. ∑ k = 1

6

k = 6(6 + 1)

— 2

= 21

17. The total number of apples in the stack is:

∑ i = 1

9

i 2 = 9(9 + 1)(2 ⋅ 9 + 1)

—— 6

= 9(10)(19)

— 6

= 285

There are 285 apples in the stack.

8.1 Exercises (pp. 414–416)

Vocabulary and Core Concept Check

1. Another name for summation notation is sigma notation.

2. In a sequence, the numbers are called terms of the sequence.

3. A sequence is an ordered list of numbers and a series is the

sum of the terms of a sequence.

4. ∑ i = 1

6

i 2 = 6(6 + 1)(2 ⋅ 6 + 1)

—— 6

= (7)(13)

= 91

∑ i = 0

5

i 2 does not belong with the other three because the others

are equal to 91.

Monitoring Progress and Modeling with Mathematics

5. a1 = 1 + 2 = 3 6. a1 = 6 − 1 = 5

a2 = 2 + 2 = 4 a2 = 6 − 2 = 4

a3 = 3 + 2 = 5 a3 = 6 − 3 = 3

a4 = 4 + 2 = 6 a4 = 6 − 4 = 2

a5 = 5 + 2 = 7 a5 = 6 − 5 = 1

a6 = 6 + 2 = 8 a6 = 6 − 6 = 0


Chapter 8

7. a1 = 12 = 1 8. f (1) = 13 + 2 = 3

a2 = 22 = 4 f (2) = 23 + 2 = 10

a3 = 32 = 9 f (3) = 33 + 2 = 29

a4 = 42 = 16 f (4) = 43 + 2 = 66

a5 = 52 = 25 f (5) = 53 + 2 = 127

a6 = 62 = 36 f (6) = 63 + 2 = 218

9. f (1) = 41 − 1 = 1 10. a1 = −12 = −1

f (2) = 42 − 1 = 4 a2 = −22 = −4

f (3) = 43 − 1 = 16 a3 = −32 = −9

f (4) = 44 − 1 = 64 a4 = −42 = −16

f (5) = 45 − 1 = 256 a5 = −52 = −25

f (6) = 46 − 1 = 1024 a6 = −62 = −36

11. a1 = 12 − 5 = −4 12. a1 = (1 + 3)2 = 16

a2 = 22 − 5 = −1 a2 = (2 + 3)2 = 25

a3 = 32 − 5 = 4 a3 = (3 + 3)2 = 36

a4 = 42 − 5 = 11 a4 = (4 + 3)2 = 49

a5 = 52 − 5 = 20 a5 = (5 + 3)2 = 64

a6 = 62 − 5 = 31 a6 = (6 + 3)2 = 81

13. f (1) = 2(1)

— 1 + 2

= 2 —

3 14. f (1) =

1 —

2(1) − 1 = 1

f (2) = 2(2)

— 2 + 2

= 1 f (2) = 2 —

2(2) − 1 = 2 —

3

f (3) = 2(3)

— 3 + 2

= 6 —

5 f (3) =

3 —

2(3) − 1 = 3 —

5

f (4) = 2(4) —

4 + 2 =

4 —

3 f (4) =

4 —

2(4) − 1 = 4 —

7

f (5) = 2(5)

— 5 + 2

= 10

— 7 f (5) =

5 —

2(5) − 1 = 5 —

9

f (6) = 2(6)

— 6 + 2

= 3 —

2 f (6) =

6 —

2(6) − 1 = 6 —

11

15. You can write the terms as 5(1) − 4, 5(2) − 4, 5(3) − 4,

5(4) − 4, . . . .

The next term is a5 = 5(5) − 4 = 21.

A rule for the nth term is an = 5n − 4.

16. You can write the terms as 21 − 1, 22 − 1, 23 − 1, 24 − 1, . . . .

The next term is a5 = 25 − 1 = 16.

A rule for the nth term is an = 2n − 1.

17. You can write the terms as 0.7(1) + 2.4, 0.7(2) + 2.4,

0.7(3) + 2.4, 0.7(4) + 2.4, . . . .

The next term is a5 = 0.7(5) + 2.4 = 5.9.

A rule for the nth term is an = 0.7n + 2.4.

18. You can write the terms as 7.8(1) + 1.2, 7.8(2) + 1.2,

7.8(3) + 1.2, 7.8(4) + 1.2, . . . .

The next term is a5 = 7.8(5) + 1.2 = 40.2.

A rule for the nth term is an = 7.8n + 1.2.

19. You can write the terms as −1.6(1) + 7.4, −1.6(2) + 7.4,

−1.6(3) + 7.4, −1.6(4) + 7.4, −1.6(5) + 7.4, . . . .

The next term is a6 = −1.6(6) + 7.4 = −2.2.

A rule for the nth term is an = −1.6n + 7.4.

20. You can write the terms as (−1)1 4(1), (−1)2 4(2), (−1)3 4(3),

(−1)4 4(4), . . . .

The next term is a5 = (−1)5 4(5) = −20.

A rule for the nth term is an = (−1)n 4n.

21. The terms are 1 —

4 ,

2 —

4 ,

3 —

4 ,

4 —

4 , . . . . The next term is a5 =

5 —

4 .

A rule for the nth term is an = n —

4 .

22. You can write the terms as 2(1) − 1

— 10(1)

, 2(2) − 1

— 10(2)

,

2(3) − 1

— 10(3)

, 2(4) − 1

— 10(4)

, . . . .

The next term is a5 = 2(5) − 1

— 10(5)

= 9 —

50 .

A rule for the nth term is an = 2n − 1 —

10n .

23. You can write the terms as 2 —

3(1) ,

2 —

3(2) ,

2 —

3(3) ,

2 —

3(4) , . . . .

The next term is a5 = 2 —

3(5) = 2 —

15 .

A rule for the nth term is an = 2 — 3n

.

24. You can write the terms as 2(1)

— 1 + 2

, 2(2)

— 2 + 2

, 2(3)

— 3 + 2

, 2(4)

— 4 + 2

, . . . .

The next term is a5 = 2(5)

— 5 + 2

= 10 —

7 .

A rule for the nth term is an = 2n —

n + 2 .

25. You can write the terms as 13 + 1, 23 + 1, 33 + 1, 43 + 1, . . . .

The next term is a5 = 53 + 1 = 126.

A rule for the nth term is an = n3 + 1.

26. You can write the terms as 12 + 0.2, 22 + 0.2, 32 + 0.2,

42 + 0.2, . . . .

The next term is a5 = 52 + 0.2 = 25.2.

A rule for the nth term is an = n2 + 0.2.

27. The rule is D.

The number of squares in the nth fi gure is equal to the sum

of the fi rst positive n integers, which is equal to the rule

given in D.

28. The rule is C.

The number of green squares in each fi gure can be written

as 4(1), 4(2), 4(3), . . . . So, the number of squares in the nth

fi gure is 4n.


Chapter 8

29. Step 1 Make a table showing the number of people that can

be seated in the fi rst three arrangements.

Let an represent the number of people around n tables.

Tables, n 1 2 3

Number of People, an

6 = 4(1) + 2 10 = 4(2) + 2 14 = 4(3) + 2

Step 2 Write a rule for the number of people around

n tables. From the table, you can see that

an = 4n + 2.

Step 3

n

an

24

30

36

42

48

12

18

6

04 5 6 7 83210

30. Step 1 Make a table showing the salary of the employee for

the fi rst three years. Let an represent the salary in

year n.

Year, n 1 2 3

Salary, an33,000 = 33,000

+ 2400(1 − 1)

33,000 +

2400(2 − 1)

33,000 +

2400(3 − 1)

Step 2 Write a rule for the salary of the employee in year n.

From the table, you can see that

an = 33,000 + 2400(n − 1) = 2400n + 30,600.

Step 3

n

an

1 2 3 4 5 6 7 80

65250

13,050

19,575

26,100

32,625

39,150

45,675

52,200

31. Notice that the terms of the series are 3(1) + 4, 3(2) + 4,

3(3) + 4, 3(4) + 4, and 3(5) + 4. So, the terms of the series

can be written as ai = 3i + 4, where i = 1, 2, 3, 4, 5.


The summation notation for the series is ∑ i = 1

5

( 3i+ 4 ) .

32. Notice that the terms of the series are 6(1) − 1, 6(2) − 1,

6(3) − 1, 6(4) − 1, and 6(5) − 1. So, the terms of the series

can be written as ai = 6i − 1, where i = 1, 2, 3, 4, 5.



5

( 6i − 1 ) .

33. Notice that the terms of the series are 12 + 3, 22 + 3, 32 + 3,

42 + 3, . . . . So, the terms of the series can be written as

ai = i 2 + 3, where i = 1, 2, 3, . . . .


infi nity.


∞ ( i 2 + 3 ) .

34. Notice that the terms of the series are 12 − 2, 22 − 2, 32 − 2,

42 − 2, . . . . So, the terms of the series can be written as

ai = i2 − 2, where i = 1, 2, 3, . . . .


infi nity.


∞ ( i 2 − 2 ) .

35. Notice that the terms of the series are 1 —

31 ,

1 —

32 ,

1 —

33 ,

1 —

34 , . . . .

So, the terms of the series can be written as ai = 1 —

3i ,

where i = 1, 2, 3, . . . .


infi nity.


∞

1 —

3i .

36. Notice that the terms of the series are 1 —

1 + 3 ,

2 —

2 + 3 ,

3 —

3 + 3 ,

4 —

4 + 3 , . . . . So, the terms of the series can be written as

ai = i —

i + 3 , where i = 1, 2, 3, . . . .


infi nity. The summation notation for the series is ∑ i = 1

∞

i —

i + 3 .

37. Notice that the terms of the series are (−1)1(1 + 2),

(−1)2(2 + 2), (−1)3(3 + 2), (−1)4(4 + 2), (−1)5(5 + 2).

So, the terms of the series can be written as

ai = (−1)i(i + 2), where i = 1, 2, . . . , 5.



5

(−1)i (i + 2).

38. Notice that the terms of the series are (−2)1, (−2)2, (−2)3,

(−2)4, and (−2)5. So, the terms of the series can be written

as ai = (−2)i, where i = 1, 2, . . . , 5.



5

(−2)i .

39. ∑ i = 1

6

2i = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) + 2(6)

= 2[1 + 2 + 3 + 4 + 5 + 6]

= 2 ∑ i = 1

6

i

= 2 ⋅ 6(6 + 1) —

2

= 6(7)

= 42


Chapter 8

40. ∑ i = 1

5

7i = 7(1) + 7(2) + 7(3) + 7(4) + 7(5)

= 7(1 + 2 + 3 + 4 + 5)

= 7(15)

= 105

41. ∑ n = 0

4

n3 = 03 + 13 + 23 + 33 + 43

= 1 + 8 + 27 + 64

= 100

42. ∑ k = 1

4

3k 2 = 3(1)2 + 3(2)2 + 3(3)2 + 3(4)2

= 3(1 + 4 + 9 + 16)

= 3(30)

= 90

43. ∑ k = 3

6

(5k − 2) = [ 5(3) − 2 ] + [ 5(4) − 2 ] + [ 5(5) − 2 ] +

[ 5(6) − 2 ]

= 13 + 18 + 23 + 28

= 82

44. ∑ n = 1

5

(n2 − 1) = (12 − 1) + (22 − 1) + (32 − 1) + (42 − 1) + (52 − 1)

= 0 + 3 + 8 + 15 + 24

= 50

45. ∑ i = 1

8

2 — i = 2 —

2 + 2 —

3 + 2 —

4 + 2 —

5 + 2 —

6 + 2 —

7 + 2 —

8

= 2 ( 1 — 2 + 1 —

3 + 1 —

4 + 1 —

5 + 1 —

6 + 1 —

7 + 1 —

8 )

= 2 ( 420 —

840 +

280 —

840 +

210 —

840 + 168

— 840

+ 140

— 840

+ 120

— 840

+ 105

— 840

) = 2 ( 1443

— 840

) = 481

— 140

46. ∑ k = 4

6

k —

k + 1 = 4 — 5 + 5 —

6 + 6 —

7

= 168 —

210 + 175

— 210

+ 180 —

210

= 523 —

210

47. ∑ i = 1

35

1 = 35

48. ∑ n = 1

16

n = 16(16 + 1)

— 2

= 8(17)

= 136

49. ∑ i = 10

25

i = 10 + 11 + 12 + ⋅⋅⋅ + 25

= (1 + 2 + 3 + ⋅⋅⋅ + 25) − (1 + 2 + 3 + ⋅⋅⋅ + 9)

= 25(25 + 1) —

2 − 9(9 + 1)

— 2

= 25(26) —

2 − 9(10)

— 2

= 25(13) − 9(5)

= 280

50. ∑ n = 1

18

n2 = 18(18 + 1)(2 ⋅ 18 + 1)

—— 6

= 3(19)(37)

= 2109

51. The error in fi nding the sum is that there are only 5 terms but

there should be 10 terms.

∑ n = 1

10

(3n − 5)

= −2 + 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 = 115

52. The error in fi nding the sum is that the index of the formula

begins at 1. So, the term i = 1 should be subtracted from

the sum.

∑ i = 2

4

i 2 = ∑ i = 1

4

i − 1 = 4(4 + 1) ( 2 ⋅ 4 + 1 )

—— 6 − 1

= 180

— 6 − 1

= 30 − 1

53. a. You save n pennies on day n. So, the terms of the series

can be written as ai = i, where i = 1, 2, 3, . . ., 100.

∑ i = 1

100

i = 100(100 + 1) ——

2

= 50(101)

= 5050

So, you saved 5050 pennies after 100 days. Therefore,

the money you saved after 100 days is 5050 × $0.01 =

$50.50.

b. To save $500, you need to save 500 × 100 = 50,000

pennies. To fi nd the number of days d, use the formula for

the sum of the fi rst n positive integers.

∑ n = 1

d

n = 50,000

d(d + 1)

— 2 = 50,000

d(d + 1) = 100,000

d 2 + d − 100,000 = 0

d = 1 — 2 ( −1 ± √

— 400,001 )

Using the positive solution, d = 1 —

2 ( √—

400,001 − 1 ) ≈

315.73. So, it will take 316 days for you to save at

least $500.


Chapter 8

54. The fi rst week you do 25 + 10(1 − 1) push-ups, the second

week you do 25 + 10(2 − 1) push-ups, and so on. So, the

nth term is 25 + 10(n − 1) = 15 + 10n, the number of

push-ups in the nth week. So, in the ninth week, n = 9, and

the number of push-ups is 15 + 10 ⋅ 9 = 105.

55. The first layer has 1 ball, the second layer has 1 + 2 =

3 balls, the third layer has 1 + 2 + 3 = 6 balls, the fourth

layer has 1 + 2 + 3 + 4 = 10 balls, and the fifth layer has

1 + 2 + 3 + 4 + 5 = 15 balls.

So, the number of soccer balls in layer n is

an = 1 + 2 + 3 + . . . + n = n(n + 1)

— 2 .

n

an

16

20

24

8

12

4

04 5 63210

56. From the diagram, each section of the array represents a

number in the series and the total number of circles form a

square grid with side n, so the sum is the area of the square. So,

1 + 3 + 5 + 7 + 9 + . . . + (2n − 1) = n • n

= n2.

57. Your friend is correct. To calculate the sum you can use the

formula for the sum of the fi rst n positive integers and

subtract 3 from the sum to evaluate ∑ i = 3

1659

i .

∑ i = 3

1659

i = ∑ i = 1

1659

i − (1 + 2)

= ∑ i = 1

1659

i − 3

58. a. The fi rst fi ve terms of the sequence are as follows.

a3 = 180(3 − 2)

— 3 = 60

a4 = 180(4 − 2)

— 4 = 90

a5 = 180(5 − 2)

— 5 = 108

a6 = 180(6 − 2)

— 6 = 120

a7 = 180(7 − 2)

— 7 ≈ 128.6

b. Each regular n-sided polygon has n equal angle measures.

So, Tn = nan

= n 180(n − 2)

— n

= 180(n − 2).

c. A regular dodecagon has n = 12 sides. So, the sum of

the interior angle measures in the Guggenheim Museum

skylight is

T12 = 180(12 − 2)

= 1800°.

59. a. The statement is true.

∑ i = 1

n

cai = ca1 + ca2 + ca3 + . . . + can

= c ( a1 + a2 + a3 + . . . + an )

= c ∑ i = 1

n

ai

b. The statement is true.

∑ i = 1

n

(ai + bi) = ( a1 + b1 ) + ( a2 + b2 ) + . . . + ( an + bn )

= ( a1 + a2 + . . . + an ) + ( b1 + b2 + . . . + bn )

= ∑ i = 1

n

ai + ∑ i = 1

n

bi

c. The statement is false.

Sample answer:

∑ i = 1

2

(i)(i) = (1)(1) + (2)(2)

= 1 + 4

= 5

( ∑ i = 1

2

i ) ( ∑ i = 1

2

i ) = (1 + 2)(1 + 2)

= (3)(3)

= 9

d. The statement is false.

Sample answer:

∑ i = 1

2

i2 = 12 + 22

= 5

( ∑ i = 1

2

i ) 2

= (1 + 2)2

= 9

60. A formula for the sum of the cubes of the fi rst n positive

integers is

∑ i = 1

n

i 3 = n2(n + 1)2

— 4 .

61. a. The terms of the series are a1 = 1 = 21 − 1,

a 2 = 3 = 22 − 1, a3 = 7 = 23 − 1, a4 = 15 = 24 − 1,

a5 = 31 = 25 − 1.

So, the nth term of the sequence is an = 2n − 1.

b. To move 6 rings, n = 6. So, the minimum number of

moves required is a6 = 26 − 1 = 63.

To move 7 rings, n = 7. So, the minimum number of


To move 8 rings, n = 8. So, the minimum number of



Chapter 8

Maintaining Mathematical Profi ciency

62. 2x − y − 3z = 6

x + y + 4z = −1

3x + z = 5

3x + z = 5

−3x + 2z = −8

3z = −3

z = −1

3x + (−1) = 5

3x = 6

x = 2

x + y + 4z = −1

2 + y + 4(−1) = −1

y = 1

So, the solution is (2, 1, −1).

Check 2(2) − (1) − 3(−1) = 6 ✓

(2) + (1) + 4(−1) = −1 ✓

3(2) − 2 (−1) = 8 ✓

63. 2x − 2y + z = 5

− 2x + 3y + 2z = −1

y + 3z = 4

−2x + 3y + 2z = −1

2x − 8y + 10z = 8

−5y + 12z = 7

5y + 15z = 20

−5y + 12z = 7

27z = 27

z = 1

5y + 15z = 20

5y + 15(1) = 20

5y = 5

y = 1

2x − 2y + z = 5

2x − 2(1) + (1) = 5

2x = 6

x = 3

So, the solution is (3, 1, 1).

Check 2(3) − 2(1) + (1) = 5 ✓

−2(3) + 3(1) + 2(1) = −1 ✓

(3) − 4(1) + 5(1) = 4 ✓

64. 2x − 3y + z = 4

− 2x + 4z = −2

−3y + 5z = 2

−3y + 5z = 2

3y + 3z = 6

8z = 8

z = 1

y + z = 2

y + 1 = 2

y = 1

x − 2z = 1

x − 2(1) = 1

x = 3

So, the solution is (3, 1, 1).

Check 2(3) − 3(1) + (1) = 4 ✓

(3) − 2(1) = 1 ✓

(1) + (1) = 2 ✓


1. a. The graph does not show an arithmetic sequence.

b. The graph does not show an arithmetic sequence.

c. The graph shows an arithmetic sequence. The nth term of

the sequence is an = 0.5n + 5.

Using a spreadsheet, the sum of the fi rst 20 terms is 205.

The graph of an arithmetic sequence is linear.

d. The graph shows an arithmetic sequence. The nth term of

the sequence is an = −2.5n + 15.

Using a spreadsheet, the sum of the fi rst 20 terms is −225.

The graph of an arithmetic sequence is linear.

2. Sample answer: Use each number twice to create 100 pairs

that sum to 101. A formula for the sum of the fi rst n terms of

an arithmetic sequence is Sn = n ( a1+ an )

— 2 .

For Exploration 1(c), Sn = 20(5.5 + 15)

—— 2 = 205

For Exploration 1(d), Sn = 20(12.5 − 35)

—— 2 = −225

So, the results are the same.

3. The graph of an arithmetic sequence is linear.

4. a. The number of terms in the sequence is 301 − 1

— 3 + 1 = 101.

So, the sum of the terms is 101

— 2 (1 + 301) = 15,251.

b. The number of terms in the sequence is 1000.


— 2 (1 + 1000) = 500,500.


Chapter 8

c. The number of terms in the sequence is 800 − 2

— 2 + 1 = 400.


— 2 (2 + 800) = 160,400.

8.2 Monitoring Progress (pp. 418−421)

1. The sequence is arithmetic because the difference between

the consecutive terms is 3.

2. The sequence is arithmetic because the difference between

the consecutive terms is −6.

3. The sequence is not arithmetic because the differences

between the consecutive terms is not constant.

4. The sequence is arithmetic with fi rst term a1 = 7, and common

difference d = 11 − 7 = 4. So, a rule for the nth term is

an = a1 + (n − 1)d

= 7 + (n − 1)4

= 4n + 3.

The 15th term is a15 = 4 ⋅ 15 + 3 = 63.

5. Step 1 Use the general rule to fi nd the fi rst term.

an = a1 + (n − 1)d

a11 = a1 + (11 − 1)d

50 = a1 + (10)7

−20 = a1

Step 2 Write a rule for the nth term.

an = a1 + (n − 1)d

= −20 + (n − 1)7

= 7n − 27

Step 3 Use the rule to create a table of values for the

sequence. Then plot the points.

n 1 2 3 4 5 6

an −20 −13 −6 1 8 15

n

an

10

15

5

−10

−5

−15

−20

−25

4 5 6 7 831 2

6. Step 1 Write a system of equations using an = a1 + (n − 1)d.

Substitute 16 for n to write Equation 1 and 7 for n to

write Equation 2.

a16 = a1 + (16 − 1)d ⇒ 26 = a1 + 15d

a7 = a1 + (7 − 1)d ⇒ 71 = a1 + 6d

45 = −9d

Step 2 Solve the system.

−5 = d

26 = a1 + 15(−5)

101 = a1

Step 3 Write a rule for an.

an = a1 + (n − 1)d

= 101 + (n − 1)(−5)

= −5n + 106

Step 4 n 1 2 3 4 5 6

an 101 96 91 86 81 76

n

an

88

94

106

100

76

82

70

04 5 6 73210

7. Step 1 Find the fi rst and last terms.

a1 = 9(1) = 9

a10 = 9(10) = 90

Step 2 Find the sum.

S10 = 10 ( a1 + a10 —

2 )

= 10 ( 9 + 90 —

2 )

= 495


a1 = 7(1) + 2 = 9

a12 = 7(12) + 2 = 86


S12 = 12 ( a1 + a12 —

2 )

= 12 ( 9 + 86 —

2 )

= 570


a1 = −4(1) + 6 = 2

a20 = −4(20) + 6 = −74


S20 = 20 ( a1 + a20 —

2 )

= 20 ( 2 − 74 —

2 )

= −720


Chapter 8

10. Find the sum of an arithmetic series with fi rst term a1 = 3

and last term a8 = 3(8) = 24.

S8 = 8 ( a1 + a8 —

2 )

= 8 ( 3 + 24 —

2 )

= 108

So, you need 108 cards.

8.2 Exercises (pp. 422−424)


1. The constant difference between consecutive terms of an

arithmetic sequence is called the common difference.

2. The statement “What sequence has an nth term of

an = 2n + 1?” is different than the other three.

The fi rst three statements have the same answer

an = 2n − 1, or the sequence 1, 3, 5, 7, . . . .

The last statement has the answer 3, 5, 7, . . . .


3. The sequence is arithmetic because the consecutive terms

have a common difference of −2.


have a common difference of −6.

5. The sequence is not arithmetic because the differences of

consecutive terms are not constant.








have a common difference of 1 —

4 .


have a common difference of 1 —

3 .

11. a. a1 = −3 and d = −6

A rule for the nth term is

an = a1 + (n − 1)d

= −3 + (n − 1)(−6)

= −6n + 3.

b. a1 = 7 and d = 5

A rule for the nth term is

an = a1 + (n − 1)d

= 7 + (n − 1)5

= 5n + 2.

12. When d > 0, the terms of the sequence increase. When d < 0,

the terms of the sequence decrease.

13. The sequence is arithmetic with fi rst term a1 = 12, and

common difference d = 20 − 12 = 8.

So, a rule for the nth term is

an = a1 + (n − 1)d

= 12 + (n − 1)8

= 8n + 4.

The 20th term is a20 = 8(20) + 4 = 164.


common difference d = 12 − 7 = 5.


an = a1 + (n − 1)d

= 7 + (n − 1)5

= 5n + 2.

The 20th term is a20 = 5(20) + 2 = 102.


common difference d = 48 − 51 = −3.


an = a1 + (n − 1)d

= 51 + (n − 1)(−3)

= −3n + 54.

The 20th term is a20 = −3(20) + 54 = −6.


common difference d = 79 − 86 = −7.


an = a1 + (n − 1)d

= 86 + (n − 1)(−7)

= −7n + 93.

The 20th term is a20 = −7(20) + 93 = −47.

17. The sequence is arithmetic with fi rst term a1 = −1 and

common difference d = − 1 — 3 + 1 =

2 —

3 . So, a rule for the

nth term is

an = a1 + (n − 1)d

= −1 + (n − 1) ( 2 — 3 )

= 2 —

3 n −

5 —

3 .

The 20th term is a20 = 2 —

3 (20) −

5 —

3 =

35 —

3 .


common difference d = − 5 — 4 + 2 =

3 —

4 . So, a rule for the

nth term is

an = a1 + (n − 1)d

= −2 + (n − 1) ( 3 — 4 )

= 3 —

4 n −

11 —

4 .

The 20th term is a20 = 3 —

4 (20) −

11 —

4 =

49 —

4 .


Chapter 8

19. The sequence is arithmetic with fi rst term a1 = 2.3 and

common difference d = 1.5 − 2.3 = −0.8. So, a rule for the

nth term is

an = a1 + (n − 1)d

= 2.3 + (n − 1)(−0.8)

= −0.8n + 3.1.

The 20th term is a20 = −0.8(20) + 3.1 = −12.9.

20. The sequence is arithmetic with fi rst term a1 = 11.7 and

common difference d = 10.8 − 11.7 = −0.9. So, a rule for

the nth term is

an = a1 + (n − 1)d

= 11.7 + (n − 1)(−0.9)

= −0.9n + 12.6.

The 20th term is a20 = −0.9(20) + 12.6 = −5.4.

21. The error is that the formula for the nth term should be

an = a1 + (n − 1)d

= 22 + (n − 1)(−13)

= −13n + 35.

22. The error is that the fi rst term and the common difference are

switched. A rule for the nth term is

an = 22 + (n − 1)(−13)

= −13n + 35.

23. Step 1 Find the fi rst term.

an = a1 + (n − 1)d

a11 = a1 + (11 − 1)d

43 = a1 + 10(5)

−7 = a1


an = a1 + (n − 1)d

= −7 + (n − 1)5

= 5n − 12

Step 3 n 1 2 3 4 5 6

an −7 −2 3 8 13 18

n

an

10

15

20

5

−10

−54 5 6 731


an = a1 + (n − 1)d

a13 = a1 + (13 − 1)d

42 = a1 + 12(4)

−6 = a1


an = a1 + (n − 1)d

= −6 + (n − 1)4

= 4n − 10

Step 3 n 1 2 3 4 5 6

an −6 −2 2 6 10 14

n

an

10

15

20

5

−10

−54 5 6 731


an = a1 + (n − 1)d

a20 = a1 + (20 − 1)d

−27 = a1 + 19(−2)

11 = a1


an = a1 + (n − 1)d

= 11 + (n − 1)(−2)

= −2n + 13

Step 3 n 1 2 3 4 5 6

an 11 9 7 5 3 1

n

an

8

10

12

4

6

2

04 5 63210


Chapter 8


an = a1 + (n − 1)d

a15 = a1 + (15 − 1)d

−35 = a1 + 14(−3)

7 = a1


an = a1 + (n − 1)d

= 7 + (n − 1)(−3)

= −3n + 10

Step 3 n 1 2 3 4 5 6

an 7 4 1 −2 −5 −8

n

an

4

6

8

2

−4

−6

−8

−24 5 6 7 831 2


an = a1 + (n − 1)d

a17 = a1 + (17 − 1)d

−5 = a1 + 16 ( − 1 — 2 )

3 = a1


an = a1 + (n − 1)d

= 3 + (n − 1) ( − 1 — 2 )

= − 1 — 2 n + 7 —

2

Step 3 n 1 2 3 4 5 6

an 3 2.5 2 1.5 1 0.5

n

an

4

5

6

2

3

1

04 5 63210


an = a1 + (n − 1)d

a21 = a1 + (21 − 1)d

−25 = a1 + 20 ( − 3 —

2 )

5 = a1


an = a1 + (n − 1)d

= 5 + (n − 1) ( − 3 —

2 )

= − 3 —

2 n +

13 —

2

Step 3 n 1 2 3 4 5 6

an 5 3.5 2 0.5 −1 −2.5

n

an

4

6

8

2

−4

−6

−8

−24 6 7 831 2

29. an = a1 + (n − 1)d

a8 = a1 + (8 − 1)d

−13 = a1 + 7(−8)

43 = a1

A rule for the nth term of the sequence is

an = a1 + (n − 1)d

= 43 + (n − 1)(−8)

= −8n + 51.

So, the answer is C.

30. Use a12 = 43 and d = 6 to fi nd the terms of the sequence.

n 12 11 10 9 8 7 6 5 4 3

an 43 37 31 25 19 13 7 1 −5 −11

So, another term is a3 = −11.

The answer is A.


a5 = a1 + (5 − 1)d ⇒ 41 = a1 + 4d

a10 = a1 + (10 − 1)d ⇒ 96 = a1 + 9d

−55 = −5d


11 = d

41 = a1 + 4(11)

−3 = a1


an = a1 + (n − 1)d

= −3 + (n − 1)11

= 11n − 14


Chapter 8


a7 = a1 + (7 − 1)d ⇒ 58 = a1 + 6d

a11 = a1 + (11 − 1)d ⇒ 94 = a1 + 10d

−36 = −4d


9 = d

58 = a1 + 6(9)

4 = a1


an = a1 + (n − 1)d

= 4 + (n − 1)9

= 9n − 5


a6 = a1 + (6 − 1)d ⇒ −8 = a1 + 5d

a15 = a1 + (15 − 1)d ⇒ −62 = a1 + 14d

54 = −9d


−6 = d

−8 = a1 + 5(−6)

22 = a1


an = a1 + (n − 1)d

= 22 + (n − 1)(−6)

= −6n + 28


a8 = a1 + (8 − 1)d ⇒ −15 = a1 + 7d

a17 = a1 + (17 − 1)d ⇒ −78 = a1 + 16d

63 = −9d


−7 = d

−15 = a1 + 7(−7)

34 = a1


an = a1 + (n − 1)d

= 34 + (n − 1)(−7)

= −7n + 41


a18 = a1 + (18 − 1)d ⇒ −59 = a1 + 17d

a21 = a1 + (21 − 1)d ⇒ −71 = a1 + 20d

12 = −3d


−4 = d

−59 = a1 + 17(−4)

9 = a1


an = a1 + (n − 1)d

= 9 + (n − 1)(−4)

= −4n + 13


a12 = a1 + (12 − 1)d ⇒ −38 = a1 + 11d

a19 = a1 + (19 − 1)d ⇒ −73 = a1 + 18d

35 = −7d


−5 = d

−38 = a1 + 11(−5)

17 = a1


an = a1 + (n − 1)d

= 17 + (n − 1)(−5)

= −5n + 22


a8 = a1 + (8 − 1)d ⇒ 12 = a1 + 7d

a16 = a1 + (16 − 1)d ⇒ 22 = a1 + 15d

−10 = −8d


5 — 4 = d

12 = a1 + 7 ( 5 — 4 )

13 —

4 = a1


an = a1 + (n − 1)d

= 13

— 4 + (n − 1) ( 5 —

4 )

= 5 —

4 n + 2


a12 = a1 + (12 − 1)d ⇒ 9 = a1 + 11d

a27 = a1 + (27 − 1)d ⇒ 15 = a1 + 26d

−6 = −15d


0.4 = d

9 = a1 + 11(0.4)

4.6 = a1


Chapter 8


an = a1 + (n − 1)d

= 4.6 + (n − 1)(0.4)

= 0.4n + 4.2

39. The terms of the sequence are a1 = −3(1) + 12,

a2 = −3(2) +12, a3 = −3(3) + 12, a4 = −3(4) + 12, . . . .

So, a rule for the sequence is an = −3n + 12.

40. The terms of the sequence are a1 = −5(1) + 20,

a2 = −5(2) + 20, a3 = −5(3) + 20, a4 = −5(4) + 20, . . . .

So, a rule for the sequence is an = −5n + 20.

41. The terms of the sequence are a1 = 3(1) − 7,

a2 = 3(2) − 7, a3 = 3(3) − 7, a4 = 3(4) − 7, . . . .

So, a rule for the sequence is an = 3n − 7.

42. The terms of the sequence are a1 = 7(1) − 12,

a2 = 7(2) − 12, a3 = 7(3) − 12, a4 = 7(4) − 12, . . . .

So, a rule for the sequence is an = 7n − 12.

43. The terms of the sequence are a4 = 4(4) + 9, a5 = 4(5) + 9,

a6 = 4(6) + 9, a7 = 4(7) + 9, a8 = 4(8) + 9, . . . . So, a rule

for the sequence is an = 4n + 9.

44. The terms of the sequence are a4 = 8(4) − 1, a5 = 8(5) − 1,

a6 = 8(6) − 1, a7 = 8(7) − 1, a8 = 8(8) − 1, . . . . So, a rule

for the sequence is an = 8n − 1.

45. Sample answer: The graph of an consists of discrete points,

whereas the graph of f is a continuous line.

46. The common difference doubles. Doubling two numbers

doubles the difference. If a and b are two numbers with a

difference d = b − a, then 2b − 2a = 2(b − a) = 2d.


a1 = 2(1) − 3 = −1

a20 = 2(20) − 3 = 37


S20 = 20 ( a1 + a20 —

2 )

= 20 ( −1 + 37 —

2 )

= 360


a1 = 4(1) + 7 = 11

a26 = 4(26) + 7 = 111


S26 = 26 ( a1 + a26 —

2 )

= 26 ( 11 + 111 —

2 )

= 1586


a1 = 6 − 2(1) = 4

a33 = 6 − 2(33) = −60


S33 = 33 ( a1 + a33 —

2 )

= 33 ( 4 − 60 —

2 )

= −924


a1 = −3 − 4(1) = −7

a31 = −3 − 4(31) = −127


S31 = 31 ( a1 + a31 —

2 )

= 31 ( −7 − 127 —

2 )

= −2077


a1 = −2.3 + 0.1(1) = −2.2

a41 = −2.3 + 0.1(41) = 1.8


S41 = 41 ( a1 + a41 —

2 )

= 41 ( −2.2 + 1.8 —

2 )

= −8.2


a1 = −4.1 + 0.4(1) = −3.7

a39 = −4.1 + 0.4(39) = 11.5


S39 = 39 ( a1 + a39 —

2 )

= 39 ( −3.7 + 11.5 ——

2 )

= 152.1

53. The fi rst term of the sequence is a1 = 9. The common

difference is d = 2 − 9 = −7. So, the nth term of the

sequence is

an = a1 + (n − 1)d = 9 + (n − 1)(−7) = −7n + 16.

The last term of the sequence is a19 = −7(19) + 16 = −117.

The sum of the fi rst 19 terms of the sequence is

S19 = 19 ( a1 + a19 —

2 )

= 19 ( 9 − 117 —

2 )

= −1026.


Chapter 8

54. The fi rst term of the sequence is a1 = 17. The common

difference is d = 9 − 17 = −8. So, the nth term of the

sequence is

an = a1 + (n − 1)d = 17 + (n − 1)(−8) = −8n + 25.

The last term of the sequence is a22 = −8(22) + 25 = −151.

The sum of the fi rst 22 terms of the sequence is

S22 = 22 ( a1 + a22 —

2 )

= 22 ( 17 − 151 —

2 )

= −1474.

55. a. Starting with the fi rst row, the number of band members

in the rows are 3, 5, 7, 9, . . . . These numbers form an

arithmetic sequence with fi rst term 3 and common

difference 2. So, a rule for the sequence is:

an = a1 + (n − 1)d

= 3 + (n − 1)2

= 2n + 1

So, the number of band members in the nth row is 2n + 1.

b. Find the sum of an arithmetic series with a1 = 3 and

a7 = 2(7) + 1 = 15.

S7 = 7 ( a1 + a7 —

2 ) = 7 ( 3 + 15

— 2 ) = 63

So, there are 63 band members in a formation with seven

rows.

56. a. Starting with the fi rst ring, the number of cells in the rings

are 6, 12, . . . . These numbers form an arithmetic sequence

with fi rst term 6 and common difference 6. So, a rule for

the sequence is

an = a1 + (n − 1)d

= 6 + (n − 1)6

= 6n.

So, the number of cells in the nth ring is 6n.

b. Find the sum of an arithmetic series with a1 = 6 and

a9 = 6(9) = 54.

S9 = 9 ( a1 + a9 —

2 ) = 9 ( 6 + 54

— 2 ) = 270.

Because the bees start with a single cell and form rings

around this cell, the number of cells in the honeycomb

after the ninth ring is formed is 270 + 1 = 271.

57. The total area of the rectangles around each square are

4(2) = 8, 4(4) = 8(2), 4(6) = 8(3), and 4(8) = 8(4).

So, using summation notation, the sum of the areas of all the

strips is

S = 1 + ∑ i = 1

4

8i

= 1 + 8 + 8(2) + 8(3) + 8(4)

= 1 + 8(1 + 2 + 3 + 4)

= 1 + 8(10)

= 81.

The area is 81 square feet.

58. Graph D represents an arithmetic sequence because the

points lie on a straight line and the domain is discrete.

59. Your friend is not correct.

It is not always true that doubling the common difference

doubles the sum of a series. For example,

1 + 2 + 3 = 6, for common difference 1

whereas, 1 + 3 + 5 = 9, for common difference 2.

60. The fi rst 10 prime numbers are

3, 3 + 4 = 7, 3 + 2(4) = 11, 3 + 4(4) = 19, 3 + 5(4) = 23,

3 + 7(4) = 31, 3 + 10(4) = 43, 3 + 11(4) = 47, 3 + 14(4) = 59,

3 + 16(4) = 67.

61. The positive odd integers form an arithmetic sequence with

fi rst term a1 = 1 and common difference d = 2. So, the nth

term is

an = a1 + (n − 1)d

= 1 + (n − 1)2

= 2n − 1.

So, the last term of the sequence is a150 = 2(150) − 1 = 299.

The sum of the positive odd integers less than 300 is

S150 = 150 ( a1 + a150 —

2 )

= 150 ( 1 + 299 —

2 )

= 22,500

62. a. ∑ i = 1

n

(3i + 5) = 544

n — 2 (8 + 3n + 5) = 544

3n2 + 13n − 1088 = 0

(n − 17)(3n + 64) = 0

n − 17 = 0 or 3n + 64 = 0

n = 17 n = − 64

— 3

Because n must be a positive integer, n = 17.


Chapter 8

b. ∑ i = 1

n

(−4i − 1) = −1127

n —

2 (−5 − 4n − 1) = −1127

n —

2 (−4n − 6) + 1127 = 0

2n2 + 3n − 1127 = 0

(n − 23)(2n + 49) = 0

n − 23 = 0 or 2n + 49 = 0

n = 23 n = − 49

— 2


c. ∑ i = 5

n

(7 + 12i) = 455

∑ i = 1

n

(7 + 12i) − 19 − 31 − 43 − 55 = 455

∑ i = 1

n

(7 + 12i) = 603

n — 2 (19 + 7 + 12n) = 603

12n2 + 26n —

2 = 603

6n2 + 13n − 603 = 0

(n − 9)(6n + 67) = 0

n − 9 = 0 or 6n + 67 = 0

n = 9 n = − 67

— 6


d. ∑ i = 3

n

(−3 − 4i) = −507

∑ i = 1

n

(−3 − 4i) − (−7) − (−11) = −507

∑ i = 1

n

(−3 − 4i) = −525

n — 2 (−7 − 3 − 4n) = −525

4n2 + 10n —

2 = 525

2n2 + 5n − 525 = 0

(n − 15)(2n + 35) = 0

n − 15 = 0 or 2n + 35 = 0

n = 15 n = − 35

— 2


63. Suppose ai represents the number of seats in row i.

Then an = x, and y = ∑ i = 1

n

ai is the sum of an arithmetic sequence.

So, y = n —

2 (a1 + an)

y = n —

2 (a1 + x)

2y — n

= a1 + x

2y

— n − x = a1

So, there are 2y

— n − x seats in the front row of the theater.

64. Because the sequence is arithmetic,

x − (3 − x) = (1 − 3x) − x

2x − 3 = 1 − 4x

6x = 4

x = 2 —

3

So, the terms of the sequence are 7 —

3 ,

2 —

3 , −1, . . . , with a

common difference − 5 —

3 . Therefore, the next term is

−1 − 5 — 3 = − 8 —

3 .

65. Suppose ai represents the portion of a hekat that ith man

should receive.

So, ∑ i = 1

10

ai = 10

10 —

2 (a1 + a10) = 10

a1 + a10 = 2.

a10 = a1 + (10 − 1) ( 1 — 8 )

a10 − a1 = 9 —

8

a10 + a1 = 2

2a10 = 25

— 8

a10 = 25

— 16

⇒ a1 = 7 —

16 .

So, the portion of a hekat that each man should receive is

7 — 16

, 9 —

16 ,

11 —

16 ,

13 —

16 ,

15 —

16 ,

17 —

16 ,

19 —

16 ,

21 —

16 ,

23 —

16 , and

25 —

16 .


66. 7 —

71/3 = 71 − 1/3 = 72/3

67. 3−2

— 3−4 = 3−2 − (−4) = 32

68. ( 9 — 49

) 1/2

= 91/2

— 491/2

= 3 —

7

69. ( 51/2 ⋅ 51/4 ) = 51/2 + 1/4 = 53/4


Chapter 8

70. The function represents exponential growth.

x

y

8

4

6

2

−4

−6

−8

−10

8 10642−4−2−6−8−10

y = 2ex

71. The function represents exponential decay.

x

y

8 10642−4−2−6−8

2

4

6

8

10y = e−3x

72. The function represents exponential decay.

x

y

4

2

−4

−6

−8

−10

8 10642−4−2−6−8−10

y = 3e−x

73. The function represents exponential growth.

x

y

8 10642−4−2−6−8

8

10

4

6

y = e0.25x


1. a. The graph shows a geometric sequence.

A rule for the nth term of the sequence is an = 2n

— 4 .

Using a spreadsheet, the sum of the fi rst 20 terms is

524, 287.5.

The graph of a geometric sequence is exponential.

b. The graph shows a geometric sequence.

A rule for the nth term of the sequence is an = 32 ( 1 — 2 ) n.

Using a spreadsheet, the sum is about 32.

The graph of a geometric sequence is exponential.

c. The graph does not show a geometric sequence.

d. The graph does not show a geometric sequence.

2. Sn = a1 + a1r + a1r 2 + a1r 3 + ... + a1r n−1

rSn = a1r + a1r 2 + a1r 3 + a1r 4 + ... + a1r n

So, Sn − rSn = a1 − a1r n

(1 − r) Sn = a1 (1 − r n)

Sn = a1 ( 1 − r n —

1 − r ) .

The sum of the terms in Exploration 1(a) is

S20 = 1 —

2 ( 1 − 220

— 1 − 2

) = 524,287.5.

The sum of the terms in Exploration 1(b) is

S20 = 16 ( 1 − 0.520

— 1 − 0.5

) ≈ 32.

So, the results are the same as in the spreadsheet.

3. The graph of a geometric sequence is exponential.

4. a. The nth term of the sequence is an = 2n−1. So, the last

term is a14 = 213 = 8192.

The sum of the sequence is

S14 = a1 ( 1 − r 14

— 1 − r

) = 1 ( 1 − 214

— 1 − 2

) = 16,383.

b. The nth term of the sequence is an = (0.1)n. So, the last

term is a10 = (0.1)10 = ( 1 — 10

) 10

= 10−10. The sum of the

sequence is

S10 = a1 ( 1 − r 10

— 1 − r

) = 0.1 ( 1 − (0.1)10

— 1 − 0.1

) = 0.1111111111.


1. a2

— a1 =

9 —

27 =

1 —

3 ,

a3 — a2 =

3 —

9 =

1 —

3 ,

a4 — a3 =

1 —

3 ,

a5 — a4 =

1 —

3

— 1 =

1 —

3 .

Each ratio is 1 —

3 , so the sequence is geometric.

2. a2

— a1 =

6 —

2 = 3,

a3 — a2 =

24 —

6 = 4.

Because the ratios are not constant, the sequence is not

geometric.

3. a2

— a1 =

2 —

−1 = −2,

a3 — a2 =

−4 —

2 = −2,

a4 — a3 =

8 —

−4 = −2,

a5

— a4 =

−16 —

8 = −2

Each ratio is −2, so the sequence is geometric.


Chapter 8

4. The sequence is geometric with fi rst term a1 = 3 and

common ratio r = 15

— 3 = 5. So, a rule for the nth term is

an = a1 r n−1

= 3(5)n−1.

The ninth term is

a 9 = 3(5)9−1 = 1,171,875.

5. Step 1 an = a1r n−1

a6 = a1(r)6−1

−96 = a1(−2)5

3 = a1

Step 2 an = a1r n−1

= 3(−2)n−1

So, a rule for the nth term is an = 3(−2)n−1.

Step 3 n 1 2 3 4 5 6

an 3 −6 12 −24 48 −96

n

an

20

40

−40

−20

−60

−80

−100

4 5 6 731

6. Step 1 Write a system of equations using an = a1r n−1.

a2 = a1r 2−1 ⇒ 12 = a1r

a4 = a1r 4−1 ⇒ 3 = a1r 3


12

— r = a1

3 = 12

— r r 3

3 = 12r 2

1 —

4 = r2

± √—

1 —

4 = r

± 1 —

2 = r

12 = a1r

12 = a1 ( ± 1 —

2 )

±24 = a1

Step 3 There are two possible rules for the nth term. One

rule is an = a1r n−1 = 24 ( 1 — 2 )

n−1

, and the other is

an = −24 ( − 1 — 2 )

n−1

.

Step 4 an = 24 ( 1 — 2 )

n−1

n 1 2 3 4 5 6

an 24 12 6 3 3 —

2

3 —

4

n

an

10

5

0

15

25

20

30

4 5 6 7310 2

an = −24 ( − 1 — 2 )

n−1

n 1 2 3 4 5 6

an −24 12 −6 3 − 3 — 2

3 —

4

n

an

10

−20

−10

4 62

7. Step 1 Find the fi rst term and the common ratio.

a1 = 51−1 = 50 = 1

r = 5


S8 = a1 ( 1 − r 8 —

1 − r )

= 1 ( 1 − 58

— 1 − 5

) = 97,656


a1 = 6(−2)1−1 = 6

r = −2


S12 = a1 ( 1 − r12

— 1 − r

) = 6 ( 1 − (−2)12

— 1 − (−2)

) = −8190


Chapter 8


a1 = −16(0.5)1−1 = −160 = −16

r = 0.5


S7 = a1 ( 1 − r7

— 1 − r

) = −16 ( 1 − (0.5)7

— 1 − 0.5

) = −31.75

10. If the annual interest rate is 5%, then the monthly interest

rate is i = 0.05

— 12

≈ 0.0042.

The monthly payment is M ≈ 20,000

——

∑ k = 1

60

( 1 —

1 + 0.0042 ) k

≈ 377.42.

So, the monthly payment decreases to $377.42.



1. The constant ratio of consecutive terms in a geometric

sequence is called the common ratio.

2. If the graph of the sequence is exponential, then the sequence

is geometric.

3. The nth term of a geometric sequence has the form

an = a1r n−1.

4. The sum of the fi rst n terms of a geometric series is

Sn = a1 ( 1 − r n —

1 − r ) .


5. a2

— a1 =

48 —

96 =

1 —

2 ,

a3 — a2 =

24 —

48 =

1 —

2 ,

a4 — a3 =

12 —

24 =

1 —

2 ,

a5 — a4 =

6 —

12 =

1 —

2

Each ratio is 1 —


6. a2

— a1 =

243 —

729 =

1 —

3 ,

a3 — a2 =

81 —

243 =

1 —

3 ,

a4 — a3 =

27 —

81 =

1 —

3 ,

a5 — a4 =

9 —

27 =

1 —

3

Each ratio is 1 —


7. a2

— a1 =

4 —

2 = 2,

a3 — a2 =

6 —

4 =

3 —

2

The ratios are not constant, so the sequence is not geometric.

8. a2

— a1 =

20 —

5 = 4,

a3 — a2 =

35 —

20 =

7 —

4


9. a2

— a1 =

3.2 —

0.2 = 16,

a3 — a2 =

−12.8 —

3.2 = −4


10. a2

— a1 =

−1.5 —

0.3 = −5,

a3 — a2 =

7.5 —

−1.5 = −5,

a4 — a3 =

−37.5 —

7.5 = −5,

a5

— a4 =

187.5 —

−37.5 = −5

Each ratio is −5, so the sequence is geometric.

11. a2

— a1 =

1 —

6

— 1 —

2

= 1 —

3 ,

a3 — a2 =

1 —

18

— 1 —

6

= 1 —

3 ,

a4 — a3 =

1 —

54

— 1 —

18

= 1 —

3 ,

a5

— a4 =

1 —

162

— 1 —

54

= 1 —

3

Each ratio is 1 —


12. a2

— a1 =

1 —

16

— 1 —

4

= 1 —

4 ,

a3 — a2 =

1 —

64

— 1 —

16

= 1 —

4 ,

a4 — a3 =

1 —

256

— 1 —

64

= 1 —

4 ,

a5 — a4 =

1 —

1024

—

1 —

256

= 1 —

4

Each ratio is 1 —


13. a. a1 = −3 and r = 5. So, a rule for the geometric sequence

is an = a1r n−1 = −3(5) n−1.

b. a1 = 72 and r = 1 —

3 . So, a rule for the geometric sequence

is an = a1r n−1 = 72 ( 1 — 3 )

n−1

.

14. When r > 1, the absolute values of the terms increase.

When 0 < r < 1, the absolute values of the terms decrease.

15. The sequence is geometric with a1 = 4 and r = 5.


an = a1r n−1

= 4(5)n−1.

The 7th term is a 7 = 4(5)7−1 = 62,500.

16. The sequence is geometric with a1 = 6 and r = 4.


an = a1r n−1

= 6(4)n−1.

The 7th term is a7 = 6(4)7−1 = 24,576.

17. The sequence is geometric with a1 = 112 and r = 1 —

2 .


an = a1rn−1

= 112 ( 1 — 2 ) n−1.

The 7th term is a7 = 112 ( 1 — 2 ) 7−1

= 7 — 4 .


Chapter 8

18. The sequence is geometric with a1 = 375 and r = 1 — 5 .


an = a1r n−1

= 375 ( 1 — 5 ) n−1

.

The 7th term is a7 = 375 ( 1 — 5 ) 7−1 = 3

— 125 .

19. The sequence is geometric with a1 = 4 and r = 3 —

2 .


an = a1r n−1

= 4 ( 3 — 2 ) n−1

.

The 7th term is a7 = 4 ( 3 — 2 ) 7−1

= 2916

— 64

= 729

— 16

.

20. The sequence is geometric with a1 = 2 and r = 3 — 4 .


an = a1r n−1

= 2 ( 3 — 4 ) n−1

.

The 7th term is a7 = 2 ( 3 — 4 ) 7−1 =

729 — 2048 .

21. The sequence is geometric with a1 = 1.3 and r = −3.


an = a1r n−1

= 1.3(−3)n−1.

The 7th term is a7 = 1.3(−3)7−1 = 947.7.

22. The sequence is geometric with a1 = 1.5 and r = −5.


an = a1r n−1

= 1.5(−5)n−1.

The 7th term is a7 = 1.5(−5)7−1 = 23,437.5.

23. Step 1 an = a1r n−1

a3 = a1r 3−1

4 = a1(2)2

1 = a1

Step 2 A rule for the nth term is an = a1r n−1

= 2n−1.

Step 3 n 1 2 3 4 5 6

an 1 2 4 8 16 32

n

an

10

5

0

15

25

20

30

4 5 6 7310 2

24. Step 1 an = a1r n−1

a3 = a1r 3−1

27 = a1(3)2

3 = a1

Step 2 A rule for the nth term is an = a1r n−1 = 3(3)n−1 = 3n.

Step 3 n 1 2 3 4 5 6

an 3 9 27 81 243 729

n

an

200

100

0

300

500

400

600

700

4 5 6 7310 2

25. Step 1 an = a1r n−1

a2 = a1r 2−1

30 = a1 ( 1 — 2 )

60 = a1

Step 2 A rule for the nth term is an = a1r n−1 = 60 ( 1 — 2 ) n−1

.

Step 3 n 1 2 3 4 5 6

an 60 30 15 15

— 2

15 —

4

15 —

8

n

an

20

10

0

30

50

40

60

70

4 5 6 7310 2

26. Step 1 an = a1r n−1

a2 = a1r 2−1

64 = a1 ( 1 — 4 )

256 = a1

Step 2 A rule for the nth term is an = a1r n−1 = 256 ( 1 — 4 ) n−1

.

Step 3 n 1 2 3 4 5 6

an 256 64 16 4 1 1 —

4

n

an

100

50

0

150

250

200

4 5 6 7310 2


Chapter 8

27. Step 1 an = a1r n−1

a4 = a1r 4−1

−192 = a1(4)3

−3 = a1

Step 2 A rule for the nth term is an = a1r n−1 = −3(4)n−1.

Step 3 n 1 2 3 4 5 6

an −3 −12 −48 −192 −768 −3072

nan

−2000

−2500

−3000

−3500

−1500

−500

−1000

4 5 6 731 2

28. Step 1 an = a1r n−1

a4 = a1r 4−1

−500 = a1(5)3

−4 = a1

Step 2 A rule for the nth term is an = a1r n−1 = −4(5)n−1.

Step 3

n 1 2 3 4 5 6

an −4 −20 −100 −500 −2500 −12,500

nan

−8000

−10,000

−12,000

−14,000

−6000

−2000

−4000

4 5 6 731 2

29. Step 1 an = a1rn−1

a5 = a1r5−1

3 = a1 ( − 1 — 3 )

4

243 = a1

Step 2 A rule for the nth term is an = a1rn−1 = 243 ( − 1 — 3 )

n−1.

Step 3 n 1 2 3 4 5 6

an 243 −81 27 −9 3 −1

n

an

−50

50

0

100

150

200

−100

5 6 731 2

30. Step 1 an = a1rn−1

a5 = a1r 5−1

1 = a1 ( − 1 — 5 )

4

625 = a1

Step 2 A rule for the nth term is an = a1rn−1 = 625 ( −

1 — 5 )

n−1.

Step 3

n 1 2 3 4 5 6

an 625 −125 25 −5 1 − 1 — 5

n

an

−100

100

0

200

300

400

500

600

−200

54 6 731 2

31. The formula should be an = a1r n−1.

The correct rule is: an = a1r n−1

a2 = a1r2−1

48 = a1(6)

8 = a1

So, an = 8(6)n−1.

32. The formula should be an = a1r n−1.

The correct rule is: an = a1r n−1

a2 = a1r 2−1

48 = a1(6)

8 = a1

So, an = 8(6)n−1.


a2 = a1r 2 −1 ⇒ 28 = a1r

a5 = a1r 5 −1 ⇒ 1792 = a1r 4

Step 2 Solve the system. 28

— r = a1

1792 = 28

— r (r 4)

1792 = 28r 3

64 = r 3

4 = r

28 = a1(4)

7 = a1

Step 3 A rule for the nth term is an = a1r n−1 = 7(4)n−1.


Chapter 8

34. Write and solve an equation using an = a1rn−1.

a4 = a1r4−1

88 = (11)r 3

8 = r 3

2 = r

A rule for the nth term is an = a1rn−1 = 11(2)n−1.

35. Write and solve an equation using an = a1rn−1.

a5 = a1r5−1 ⇒ −486 = a1r

4

a1 = −6 ⇒ −486 = (−6)r4

81 = r4

±3 = r

A rule for the nth term is an = a1rn−1 = − 6(3)n−1 or

an = −6(−3)n−1.

36. Step 1 Write a system of equations using an = a1rn−1.

a2 = a1r2−1 ⇒ −10 = a1r

a6 = a1r6−1 ⇒ −6250 = a1r5

Step 2 Solve the system. − 10

— r = a1

−6250 = ( − 10

— r ) r5

625 = r4

±5 = r

−10 = a1(±5)

±2 = a1

Step 3 A rule for the nth term is an = a1rn−1 = −2(5)n−1

or an = 2(−5)n−1.


a2 = a1r 2−1 ⇒ 64 = a1r

a4 = a1r 4−1 ⇒ 1 = a1r3

Step 2 Solve the system. 64

— r = a1

1 = ( 64 —

r ) r3

1 = 64r2

1 —

64 = r2

± 1 —

8 = r

64 = a1 ( ± 1 — 8 )

±512 = a1

Step 3 A rule for the nth term is an = a1rn−1 = 512 ( 1 — 8

) n−1

or an = −512 ( − 1 —

8 ) n−1

.

38. Step 1 The common ratio is r = a2

— a1 =

49 —

1 = 49.

Step 2 A rule for the nth term is an = a1rn−1

= 1(49)n−1

= (49)n−1.


a2 = a1r 2−1 ⇒ −72 = a1r

a6 = a1r 6−1 ⇒ − 1 —

18 = a1r 5

Step 2 Solve the system. −72

— r = a1

− 1 —

18 = ( −72

— r ) r 5

1 —

1296 = r 4

± 1 —

6 = r

−72 = a1 ( ± 1 —

6 )

±432 = a1

Step 3 A rule for the nth term is

an = a1r n−1 = −432 ( 1 — 6 ) n−1

or an = 432 ( − 1 — 6

) n−1

.


a2 = a1r 2−1 ⇒ −48 = a1r

a5 = a1r 5−1 ⇒ 3 —

4 = a1r 4

Step 2 Solve the system. −48

— r = a1

3 —

4 = ( −48

— r ) r 4

− 1 —

64 = r 3

− 1 —

4 = r

−48 = a1 ( − 1 —

4 )

192 = a1

Step 3 A rule for the nth term is an = a1r n−1 = 192 ( − 1 —

4 ) n−1

.

41. The terms of the sequence are a1 = 4(2)0, a2 = 4(2)1,

a3 = 4(2)2, and a4 = 4(2)3. So, a rule for the nth term is

an = 4(2)n−1.

42. The terms of the sequence are a1 = 5(3)0, a2 = 5(3)1,

a3 = 5(3)2, and a4 = 5(3)3. So, a rule for the nth term is

an = 5(3)n−1.

43. The terms of the sequence are a1 = 5 ( 1 — 2 ) 0, a2 = 5 ( 1 —

2 ) 1,

a3 = 5 ( 1 — 2 ) 2, and a4 = 5 ( 1 —

2 ) 3. So, a rule for the nth term is

an = 5 ( 1 — 2 ) n−1

.

44. The terms of the sequence are a1 = 48 ( 1 — 4 ) 0, a2 = 48 ( 1 —

4 ) 1,

a3 = 48 ( 1 — 4 ) 2, and a4 = 48 ( 1 —

4 ) 3. So, a rule for the nth term is

an = 48 ( 1 — 4 ) n−1

.


Chapter 8

45. The terms of the sequence are a2 = 6(−2)1, a3 = 6(−2)2,

a4 = 6(−2)3, …. So, a rule for the nth term is

an = 6(−2)n−1.

46. The terms of the sequence are a2 = 7(−3)1, a3 = 7(−3)2,

a4 = 7(−3)3, …. So, a rule for the nth term is

an = 7(−3)n−1.


a1 = 6(7)0 = 6

r = 7


S9 = a1 ( 1 − r 9 —

1 − r )

= 6 ( 1 − 79

— 1 − 7

) = 40,353,606


a1 = 7(4)1−1 = 7(4)0 = 7

r = 4


S10 = a1 ( 1 − r 10

— 1 − r

) = 7 ( 1 − 410

— 1 − 4

) = 2,446,675


a1 = 4 ( 3 — 4 ) 0 = 4

r = 3 —

4


S10 = a1 ( 1 − r 10

— 1 − r

)

= 4 ( 1 − ( 3 — 4 ) 10

— 1 −

3 —

4

) =

989,527 —

65,536


a1 = 5 ( 1 — 3 ) 0 = 5

r = 1 —

3


S8 = a1 ( 1 − r 8 —

1 − r )

= 5 ( 1 − ( 1 — 3 ) 8 —

1 − 1 —

3

) =

15 —

2 ( 1 − ( 1 —

3 ) 8 )

= 16,400

— 2187

51. Step 1 Rewrite the series ∑ i = 0

8

8 ( − 2 —

3 ) i as 8 + ∑

i = 1

8

8 ( − 2 —

3 ) i.

Then fi nd a1 and the common ratio.

a1 = 8 ( − 2 —

3 ) 1 = −

16 —

3

r = − 2 —

3


∑ i = 0

8

8 ( − 2 —

3 ) i = 8 + ∑

i = 1

8

8 ( − 2 —

3 ) i

= 8 + S8

= 8 + a1 ( 1 − r 8 —

1 − r )

= 8 + ( − 16

— 3 ) ( 1 − ( −

2 —

3 ) 8 —

1 − ( − 2 —

3 ) )

= 8 + ( − 16

— 3 ) ( 1 − ( 2 —

3 ) 8 —

5 —

3

) =

32,312 —

6561


Chapter 8

52. Step 1 Rewrite the series ∑ i = 0

9

9 ( − 3 —

4 ) i as 9 + ∑

i = 1

9

9 ( − 3 —

4 ) i.

Then fi nd a1 and the common ratio.

a1 = 9 ( − 3 —

4 ) 1 = −

27 —

4

r = − 3 —

4


∑ i = 0

9

9 ( − 3 —

4 ) i = 9 + ∑

i = 1

9

9 ( − 3 —

4 ) i

= 9 + S9

= 9 + a1 ( 1 − r 9 —

1 − r )

= 9 + ( − 27

— 4 ) ( 1 − ( −

3 —

4 ) 9 —

1 − ( − 3 —

4 ) )

= 9 + ( − 27

— 4 ) ( 1 + ( 3 —

4 ) 9 —

7 —

4

) =

1,272,249 —

262,144

53. Step 1 Find the common ratio.

r = −48

— −12

= 4


S8 = a1 ( 1 − r 8 —

1 − r )

= −12 ( 1 − 48

— 1 − 4

) = −262,140

54. Step 1 Find the common ratio.

r = −42

— −14

= 3


S9 = a1 ( 1 − r 9 —

1 − r )

= −14 ( 1 − 39

— 1 − 3

) = −137,774

55. The graph of an consists of discrete points and the graph of f is continuous.

56. a. 1 + x + x2 + x3 + x 4 = ∑ n = 1

5

x n−1

= 1 − x5

— 1 − x

b. 3x + 6x3 + 12x5 + 24x7 = ∑ n = 1

4

( 3x ) ( 2x2 ) n−1

= 3x ( 1 − ( 2x2 ) 4 —

1 − 2x2 )

= 3x ( 1 − 16x8

— 1 − 2x2

) =

3x − 48x 9 —

1 − 2x2

57. Step 1 Substitute for L, i, and t.

L = 15,000, i = 0.04

— 12

, and t = 5(12) = 60.

So, M = 15000 ——

∑ k = 1

60

( 1 —

1 + 0.04

— 12

) k

Step 2 Use a calculator to fi nd the monthly payment.

Notice that the denominator is a geometric series

with fi rst term 300

— 301

and common ratio 300

— 301

.

So, the monthly payment is $276.25.

58. Step 1 Substitute for L, i, and t.

L = 200,000, i = 0.045

— 12

, and t = 30(12) = 360.

So, M = 200,000 ——

∑ k = 1

360

( 1 —

1 + 0.045

— 12

) k

Step 2 Use a calculator to fi nd the monthly payment.

Notice that the denominator is a geometric series

with fi rst term 800

— 803

and common ratio 800

— 803

.

So, the monthly payment is $1013.37.

59. a. The fi rst term is 32 and the common ratio is 1 —

2 . So, a rule

for the number of games played in the nth round is

an = 32 ( 1 — 2 ) n−1

, 1 ≤ n ≤ 6.

The number of games must be a whole number, so n is

between 1 and 6.

b. The total number of games is ∑ i = 1

6

32 ( 1 — 2 ) i−1

= 32 ( 1 − ( 1 — 2 ) 6 —

1 − 1 —

2

) = 63.


Chapter 8

60. a. The fi rst term is 5 and the common ratio is 10

— 5 = 2.

So, a rule for an is an = 5(2)n−1.

b. The total number of skydivers is

∑ n = 1

4

5(2)n−1

= 5 ( 1 − 24

— 1 − 2

) = 75.

61. a. The fi rst term is 1 and the common ratio is 8.

So, a rule for an is an = 8n−1.

The total number of squares removed through stage 8 is

S8 = ∑ n = 1

8

8n−1

= 1 ( 1 − 88

— 1 − 8

) = 2,396,745.

b. The fi rst term is 1 − 1 —

9 =

8 —

9 and the common ratio is

8 —

9 .

So, a rule for bn is bn = 8 —

9 ( 8 —

9 ) n−1

= ( 8 — 9 ) n.

The remaining area after stage 12 is b12 = ( 8 — 9 ) 12

≈ 0.243

square unit.

62. a. The graph is B.

Because r = 1 —

2 < 1, the graph represents exponential decay.

b. The graph is A.

Because r = 2 > 1, the graph represents exponential growth.

63. The amount of money in your account is

∑ k = 0

30

[ 2000(1 + 0.05)k ] = $141,521.58.

64. The fi rst iteration is an equilateral triangle with side

lengths of 1. The perimeter is 3(1) = 3 and the area is

√

— 3 —

4 (1)2 =

√—

3 —

4 . In the second iteration, the middle third of

each side of the triangle is removed, and three smaller

equilateral triangles are added to the fi gure. Because each

line segment has a length of 1 —

3 , and there are 12 segments,

the perimeter is 12 ( 1 — 3 ) = 4. The total area is

√

— 3 —

4 + 3 ⋅

√—

3 —

4 ( 1 —

3 ) 2 = √

— 3 —

4 +

√—

3 —

12 =

4 √—

3 —

12 =

√—

3 —

3 . Similarly, in

the third iteration, the middle third of each line segment is

removed and 12 smaller equilateral triangles are added.

Because each new line segment has a length of 1 —

9 , and there

are 48 segments, the perimeter is 48 ( 1 — 9 ) =

16 —

3 . The total area

is √

— 3 —

3 + 12 ⋅

√—

3 —

4 ( 1 —

9 ) 2 =

√—

3 —

3 +

√—

3 —

27 =

10 √—

3 —

27 . In the last

iteration, there are 192 new line segments of length 1 —

27

and 48 new triangles. So, the perimeter is 192 ( 1 — 27

) = 64

— 9 and

the total area is 10 √

— 3 —

27 + 48 ⋅

√—

3 —

4 ⋅ ( 1 —

27 ) 2 =

10 √—

3 —

27 +

4 √—

3 —

243

= 94 √

— 3 —

243 .

The perimeters 3, 4, 16

— 3 ,

64 —

9 form a geometric sequence

because the terms have a common ratio of 4 —

3 .

The areas √

— 3 —

4 ,

√—

3 —

3 ,

10 √—

3 —

27 ,

94 √—

3 —

243 do not form a geometric

sequence because the terms do not have a common ratio.


The monthly payment for Loan 1 is

165,000

——

∑ k = 1

180

( 1 —

1 + 0.03

— 12

) k = $1139.50.

So, the total amount repaid over the Loan 1 is about

180($1139.50) = $205,110.

The monthly payment for Loan 2 is

165,000

——

∑ k=1

360

( 1 —

1 + 0.04

— 12

) k = $787.74.

So, the total amount repaid over the Loan 2 is about

360($787.74) = $283,586.4.


Chapter 8

66. a. For a 1-month loan: L(1 + i) − M = 0

L(1 + i) = M

L = M —

1 + i

For a 2-month loan:

[ L(1 + i) − M ] (1 + i) − M = 0

[ L(1 + i) − M ] (1 + i) = M

L(1 + i) − M = M —

1 + i

L(1 + i) = M —

1 + i + M

L = M —

(1 + i)2 +

M —

1 + i

L = M —

1 + i +

M —

(1 + i)2

b. For a t-month loan:

L = M —

1 + i +

M —

(1 + i)2 + … +

M —

(1 + i)t

= M [ 1 —

1 + i +

1 —

(1 + i)2 + … +

1 —

(1 + i)t ] = M ∑

k = 1

t

( 1 —

1 + i ) k

So, M = L ——

∑ k = 1

t

[ ( 1 —

1 + i )

k

]

c. Because a1 = 1 —

1 + i and r =

1 —

1 + i ,

∑ k = 1

t

( 1 —

1 + i )

k

= ( 1 —

1 + i ) [ 1 − ( 1

— 1 + i

) t ——

1 − 1 —

1 + i ]

=

1 − ( 1 —

1 + i )

t

—— 1 + i − 1

= 1 − (1 + i)−t

— i .

So, M = L ——

[ 1 − (1 + i)−t ——

i ]

= L [ i ——

1 − (1 + i)−t ] .

In Exercise 57, M = 15,000 [ 0.04

— 12

——

1 − ( 1 + 0.04

— 12

) −60 ] = $276.25.

In Exercise 58, M = 200,000 [ 0.045

— 12

——

1 − ( 1 + 0.045

— 12

) −360 ]

= $1013.37.


67.

x

y

1

2

3

4

5

−2

−3

−4

−5

4 5 6 7 82−2

The domain is all real numbers except 3.

The range is all real numbers except 0.

68.

x

y

4

2

1

5

6

7

421 3 5−2−4−3 −1−5



69.

x

y

4

2

3

5

6

−3

−4

−2

41 3 5 6 7−2−3 −1



70.

x

y

1

−4

−3

−5

−6

−7

42 3−2−4−3−5−6

The domain is all real numbers except −1.

The range is all real numbers except −2.


Chapter 8

8.1–8.3 What Did You Learn? (p. 433)

1. Each table has 4 chairs and there are 2 more chairs on the ends.

2. Sample answer: You can use a graphing calculator to fi nd the

sum of the series.

3. The domain of an is discrete, so the graph is individual

points. The domain of f(x) is continuous, so the graph

is continuous.

8.1–8.3 Quiz (p. 434)

1. The pattern is 6(1) − 5, 6(2) − 5, 6(3) − 5, 6(4) − 5, ....

The next term is 6(5) − 5 = 25.

So, the nth term is an = 6n − 5.

2. The pattern is −1.5(1), 1.5(2), −1.5(3), 1.5(4), ....

The next term is −1.5(5) = −25.

So, the nth term is an = (−1)n5n.

3. The pattern is 1 —

10 + 10 ,

2 —

10(2) + 10 ,

3 —

10(3) + 10 ,

4 —

10(4) + 10 , ….

The next term is 5 —

10(5) + 10 =

5 —

60 .

So, the nth term is an = n —

10n + 10 .

4. ∑ i = 1

15

i = 15(15 + 1)

— 2

= 120

5. ∑ i = 1

8

i−1

— i = 0 +

420 —

840 +

560 —

840 + 630

— 840

+ 672

— 840

+ 700

— 840

+ 720

— 840

+ 735

— 840

= 4437

— 840

= 1479

— 280

6. ∑ i = 3

10

i2 = ∑ i = 1

10

i2 − (4 + 1)

= 10 (10 + 1)(2 ⋅ 10 + 1)

—— 6 − 5

= 385 − 5

= 380

7. The terms of the sequence are a1 = 1 —

4 , a2 =

1 —

4 ⋅ 2, a3 =

1 —

4 ⋅ 3,

a4 = 1 —

4 ⋅ 4, and a5 =

1 —

4 ⋅ 5. So, a rule for the nth term is

an = 1 —

4 n.

8. The terms of the sequence are a1 = 1 —

2 , a2 =

1 —

2 ⋅ 2, a3 =

1 —

2 ⋅ 4,

a4 = 1 —

2 ⋅ 8, and a5 =

1 —

2 ⋅ 16. So, the nth term is an =

1 —

2 (2)n−1.

9. The terms of the sequence are a1 = 3 – 2, a2 = 3 – 2 ⋅ 2,

a3 = 3 – 2 ⋅ 3, a4 = 3 – 2 ⋅ 4, and a5 = 3 – 2 ⋅ 5. So, the nth

term is an = –2n + 3.

10. The terms have a common difference of –7. So, the sequence

is arithmetic. Because a1 = 13 and d = –7, a rule for the nth

term is an = a1 + (n – 1)d = 13 + (n – 1)(–7) = –7n + 20.

So, a9 = –7(9) + 20 = –43.

11. The terms do not have a common difference or common

ratio, so the sequence is neither arithmetic nor geometric.

You can write the terms as 1 —

1 + 1 ,

1 —

2 + 1 ,

1 —

3 + 1 ,

1 —

4 + 1 , ….

So, a rule for the nth term is an = 1 —

n + 1 , and a9 =

1 —

9 + 1

= 1 —

10 .

12. The terms have a common ratio of –3. So, the sequence is

geometric. Because a1 = 1 and r = –3, a rule for the nth term is

an = a1r n – 1 = 1 ⋅ (–3)n – 1 = (–3)n – 1. So, a9 = (–3)9 – 1 = 6561.

13. a12 = a1(12 − 1)(7)

19 = a1 + (12 – 1)7

19 = a1 + 77

–58 = a1

So, the nth term of the sequence is an = a1 + (n – 1)d

= –58 + (n – 1)(7)

= 7n – 65.

ATime

(hours)

00.5

11.5

22.5

33.5

4

BDistance(miles)

0250500750

10001250150017502000

1

23456789

1011

14. a6 = a1r 6 – 1 ⇒ –50 = a1r5

a9 = a1r 9 – 1 ⇒ –6250 = a1r8

a1 = –50

— r5

–6250 = –50

— r5

(r 8)

–6250 = –50r 3

125 = r 3

5 = r

a1 = –50

— 55

= –0.016

So, the nth term of the sequence is an = a1r n–1.

= − 2 —

125 (5)n – 1

.


Chapter 8

15. ∑ n = 1

9

(3n + 5)

= 8 + 11 + 14 + 17 + 20 + 23 + 26 +29 + 32

= 180

16. ∑ k = 1

5

11(–3)k – 2 = – 11

— 3 + 11 – 33 + 99 – 297

= –223 2 —

3

17. ∑ i = 1

12

–4 ( 1 — 2 ) i + 3

= − 1 —

4 ( 1 – ( 1 —

2 )

12

— 1 –

1 —

2

) = −

4095 —

8192

18. If an represents the number of pieces of chalk in row n, then

an = 16 – n.

So, the total number of pieces of chalk in the pile is

∑ i = 1

10

(16 – n) = 10 ( 15 + 6 —

2 ) = 105.

19. a. Your salary for the nth year of employment is

an = 45,000(1 + 0.035)n – 1

= 45,000(1.035)n – 1.

b. During your fi fth year of employment, n = 5. So your

salary will be a5 = 45,000(1.035)4 = $51,638.54.

c. Your total earnings for 10 years is

∑ i = 1

10

45,000(1.035)n – 1 = 45,000 ( 1 –1.03510

— 1 – 1.035

) = $527,912.69.


1. a. 1

A B

234567891011121314

16

110.520.2530.12540.062550.031256

7891011121314

Sum15 15

0.0156250.0078130.0039060.0019530.0009770.0004880.0002440.0001220.0000611.999939

The infi nite geometric series has a fi nite sum of 2 because

as the number of terms increases, the sum approaches 2.

b. 1

A B

2345678910111213

110.33333320.11111130.03703740.01234650.0041156

789101112

Sum

0.0013720.0004570.0001520.0000510.0000170.0000061.499997

The infi nite geometric series has a fi nite sum of 1.5

because as the number of terms increases, the sum

approaches 1.5.

c. 1

A B

234567891011121314

16

111.522.2533.37545.062557.593756

7891011121314

Sum15

11.3906317.0859425.6289138.4433657.6650486.49756129.74634194.61951581.85852

The infi nite geometric series does not have a fi nite

sum because as the number of terms increases, the sum

increases at an approximately exponential rate.

d. 1

A B

234567891011121314

16

111.252

1.562531.9531342.4414153.05176

7891011121314

Sum15 15

3.814704.768375.960467.450589.3132311.6415314.5519218.1898922.73737109.68684

The infi nite geometric series does not have a fi nite

sum because as the number of terms increases, the sum

increases at an approximately exponential rate.


Chapter 8

e. 1

A B

234567891011121314

16

110.820.6430.51240.409650.327686

7891011121314

Sum15

0.262140.209720.167770.134220.107370.085900.068720.054984.82408

The infi nite geometric series has a fi nite sum of 5 because

as the number of terms increase, the sum approaches 5.

f. 1

A B

234567891011121314

16

110.920.8130.72940.656150.590496

78910111213141515

0.531440.478300.430470.387420.348680.313810.282430.254190.22877

1716 0.20589

49 49 0.2058950 50 0.0057351 51 0.005155253

Sum 9.95362

The infi nite geometric series has a fi nite sum of

10 because as the number of terms increases, the sum

approaches 10.

2. An infi nite geometric series has a fi nite sum when ∣ r ∣ < 1.

3. As n increases, r n approaches 0 because when ∣ r ∣ < 1, r n is

an exponential decay function.

So, the sum is S = a1 —

1 − r .

1a. S = 1 —

1 − 1 —

2

= 1

— 1 —

2

= 2

1b. S = 1 —

1 − 1 —

3

= 1

— 2 —

3

= 1.5

1e. S = 1 —

1 − 4 —

5

= 1

— 1 —

5

= 5

1f. S = 1 —

1 − 9 —

10

= 1

— 1 —

10

= 10

4. The sum of an infi nite geometric series is S = a1 —

1 − r .

5. a. S = a1 —

1 − r =

1 —

1 − 0.1 =

1 —

0.9 =

10 —

9

b. S = a1 —

1 − r =

2 —

1 − 2 —

3

= 2

— 1 —

3

= 6


1. Step 1 Find the partial sums.

S1 = 2 —

5 = 0.4

S2 = 2 —

5 +

4 —

25 = 0.56

S3 = 2 —

5 + 4 — 25 +

8 —

125 ≈ 0.62

S4 = 2 —

5 +

4 —

25 +

8 —

125 +

16 —

1625 ≈ 0.64

S5 = 2 —

5 +

4 —

25 +

8 —

125 +

16 —

1625 +

32 —

3125 ≈ 0.65

Step 2

n

sn

0.4

0.5

0.7

0.6

0.2

0.3

0.1

04 5 63210

From the graph, Sn appears to approach 0.67.

2. For the series ∑ n = 1

∞ ( −

1 —

2 ) n−1

, a1 = ( − 1 —

2 ) 1−1

= 1 and r = − 1 —

2 .

The sum of the series is

S = a1 —

1 − r

= 1 —

1 − ( − 1 —

2 )

= 2 —

3 .

3. For the series ∑ n = 1

∞ 3 ( 5 —

4 )

n−1

, a1 = 3 ( 5 — 4 )

1−1

= 3 and r = 5 —

4 .

Because ∣ 5 — 4 ∣ ≥ 1, the sum does not exist.

4. For the series 3 + 3 —

4 +

3 —

16 +

3 —

64 + . . . , a1 = 3 and a2 =

3 —

4 .

So, r =

3 —

4

— 3 =

1 —

4 .


S = a1 —

1 − r

= 3 —

1 − 1 —

4

= 4.


Chapter 8

5. The total distance traveled by the pendulum is given by

10 + 10(0.8) + 10(0.8)2 + . . . .

So, a1 = 10 and r = 0.8. The sum of the series is

S = a1 —

1 − r

= 10 —

1 − 0.8

= 50.

The pendulum travels a total distance of 50 inches.

6. 0.555 . . . = 0.5 + 0.05 + 0.005 + . . . .

For this series, a1 = 0.5 and r = 0.05

— 0.5

= 0.1. The sum of the

series is

S = a1 —

1 − r

= 0.5 —

1 − 0.1

= 0.5

— 0.9

= 5 —

9 .

7. 0.727272 . . . = 0.72 + 0.0072 + 0.000072 + . . .


— 0.72

= 0.01. The sum

of the series is

S = a1 —

1 − r

= 0.72 —

1 − 0.01

= 0.72

— 0.99

= 72

— 99

= 8 — 11

.

8. 0.131313 . . . = 0.13 + 0.0013 + 0.000013 + . . .


— 0.13

= 0.01. The sum

of the series is

S = a1 —

1 − r

= 0.13 —

1 − 0.01

= 0.13

— 0.99

= 13

— 99

.



1. The sum Sn of the fi rst n terms of an infi nite series is called a

partial sum.

2. The series has a sum when ∣ r ∣ < 1.



S1 = 1 —

2 = 0.5

S2 = 1 —

2 +

1 —

6 ≈ 0.67

S3 = 1 —

2 +

1 —

6 +

1 —

18 ≈ 0.72

S4 = 1 —

2 +

1 —

6 +

1 —

18 +

1 —

54 ≈ 0.74

S5 = 1 —

2 +

1 —

6 +

1 —

18 +

1 —

54 +

1 —

162 ≈ 0.75.

Step 2

n

0.4

sn

0.5

0.7

0.8

0.6

0.2

0.3

0.1

04 5 63210



S1 = 2 —

3 ≈ 0.67

S2 = 2 —

3 +

1 —

3 ≈ 1

S3 = 2 —

3 +

1 —

3 +

1 —

6 ≈ 1.17

S4 = 2 —

3 +

1 —

3 +

1 —

6 +

1 —

2 ≈ 1.25

S5 = 2 —

3 +

1 —

3 +

1 —

6 +

1 —

2 +

1 —

24 ≈ 1.29

Step 2

n

0.8

sn

1.0

1.4

1.2

0.6

0.4

0.2

04 5 63210

From the graph, Sn appears to approach 4 —

3 .


Chapter 8


S1 = 4

S2 = 4 + 12

— 5 = 6.4

S3 = 4 + 12

— 5 +

36 —

25 = 7.84

S4 = 4 + 12

— 5 +

36 —

25 +

108 —

125 ≈ 8.70

S5 = 4 + 12

— 5 +

36 —

25 +

108 —

125 +

324 —

625 ≈ 9.22

Step 2

n

8

sn

10

12

6

4

2

04 5 63210

From the graph, Sn appears to approach 10.


S1 = 2

S2 = 2 + 2 —

6 ≈ 2.33

S3 = 2 + 2 —

6 +

2 —

36 ≈ 2.39

S4 = 2 + 2 —

6 +

2 —

36 +

2 —

216 ≈ 2.40

S5 = 2 + 2 —

6 +

2 —

36 +

2 —

216 +

2 —

1296 ≈ 2.40

Step 2

n

2.0

sn

2.5

3.0

1.5

1.0

0.5

04 5 63210


7. For this series, a1 = 8 ( 1 — 5 ) 1 − 1

= 8 and r = 1 —

5 . The sum of the

series is

S = a1 —

1 − r

= 8 —

1 − 1 —

5

= 10.

8. For this series, a1 = −6 ( 3 — 2 )

1−1

= −6 and r = 3 —

2 .


9. For this series, a1 = 11

— 3 ( 3 —

8 ) 1 − 1

= 11

— 3 and r =

3 —

8 . The sum of

the series is

S = a1 —

1 − r

= 11

— 3

— 1 −

3 —

8

= 88

— 15

.

10. For this series, a1 = 2 —

5 ( 5 —

3 )

1 − 1 =

2 —

5 and r =

5 —

3 .


11. For this series, a1 = 2 and a2 = 6 —

4 . So, r =

6 —

4 —

2

= 6 —

8 . The

sum of the series is

S = a1 —

1 − r

= 2 —

1 − 6 —

8

= 8.

12. For this series, a1 = –5 and a2 = –2. So, r = −2

— −5

= 2 —

5 .


S = a1 —

1 − r

= −5

— 1 −

2 —

5

= − 25

— 3 .

13. For this series, a1 = 3 and a2 = 5 —

2 . So, r =

5 —

2 —

3

= 5 —

6 . The


S = a1 —

1 − r

= 3 —

1 − 5 —

6

= 18.

14. For this series, a1 = 1 —

2 and a2 = −

5 —

3 . So, r =

− 5 —

3

— 1 —

2

= − 10 —

3 .

Because ∣ − 10

— 3 ∣ ≥ 1, the sum does not exist.

15. The sum does not exist, because ∣ 7 — 2 ∣ > 1.


Chapter 8

16. The value of the r is

8 —

3

— 4 =

2 —

3 . Because ∣ 2 —

3 ∣ < 1, the sum

exist and the sum of the series is

S = a1 —

1 − r

= 4 —

1 − 2 —

3

= 12.

17. The total distance traveled by your cousin is given by the

infi nite series 14 + 14(0.75) + 14(0.75)2 + . . . .

For this series, a1 = 14 and r = 0.75. The sum of the series is

S = a1 —

1 − r

= 14 —

1 − 0.75

= 56.

Your cousin travels a total distance of 56 feet.

18. The total profi t made by the company is given by the infi nite

series

350,000 + 350,000 (1 – 0.12) + 350,000 (1 – 0.12)2 + . . .

For this series, a1 = 350,000 and r = 1 – 0.12 = 0.88. The


S = a1 —

1 − r

= 350,000

— 1 − 0.88

≈ $2,916,667.

The company can make a total profi t of about $2,916,667.

19. 0.222 . . . = 0.2 + 0.02 + 0.002 + . . .


— 0.2


series is

S = a1 —

1 − r

= 0.2 —

1 − 0.1

= 0.2

— 0.9

= 2 —

9 .

20. 0.444 . . . = 0.4 + 0.04 + 0.004 + . . .


— 0.4


series is

S = a1 —

1 − r

= 0.4 —

1 − 0.1

= 0.4

— 0.9

= 4 —

9 .

21. 0.161616 . . . = 0.16 + 0.0016 + 0.000016 + . . .

For this series, a1 = 0.16 and r = 0.0016 —

0.16 = 0.01. The sum

of the series is

S = a1 —

1 − r

= 0.16 —

1 − 0.01

= 0.16 —

0.99

= 16 —

99 .

22. 0.625625625 . . . = 0.625 + 0.000625 + 0.000000625 + . . .

For this series, a1 = 0.625 and r = 0.000625 —

0.625 = 0.001. The


S = a1 —

1 − r

= 0.625

— 1 − 0.001

= 0.625

— 0.999

= 625

— 999

.

23. 32.323232 . . . = 32 + 0.32 + 0.0032 + 0.000032 + . . .

For this series, a1 = 32 and r = 0.32

— 32


series is

S = a1 —

1 − r

= 32 —

1 − 0.01

= 32 —

0.99

= 3200 —

99

= 32 32

— 99

.


Chapter 8

24. 130.130130130 . . .

= 130 + 0.130 + 0.000130 + 0.000000130 + . . .

For this series, a1 = 130 and r = 0.130

— 130

= 0.001. The sum of

the series is

S = a1 —

1 − r

= 130 —

1 − 0.001

= 130

— 0.999

= 130,000

— 999

= 130 130

— 999

.

25. Sample answer: Two infi nite geometrics series are

∑ i = 1

∞ 3 ( 1 —

2 ) i − 1

and ∑ i = 1

∞ 2 ( 2 —

3 ) i − 1

.

The sums of the two series are

∑ i = 1

∞ 3 ( 1 —

2 ) i − 1

= 3 —

1 − 1 —

2

= 6

and ∑ i = 1

∞ 2 ( 2 —

3 ) i − 1

= 2 —

1 − 2 —

3

= 6.

26. The value of ∑ n = 1

∞ an is 1.5, because the partial sums appear to

approach 1.5.

27. The total amount of prize money is given by an infi nite series

500 + 500(0.9) + 500(0.9)2 + . . . .

In this series, a1 = 500 and r = 0.9. The sum of the series is

S = a1 —

1 − r =

500 —

1 − 0.9

= 5000.

So, the total amount of prize money is $5000.

28. The blue triangle has a base length of 4 and a height of 4. So,

the area of the blue triangle is A = 1 — 2 bh =

1 —

2 ⋅ 4 ⋅ 4 = 8.

Using Archimedes’ result, the area of the region is 4 —

3 (8) =

32 —

3 .

Next, calculate the area of the red triangles. One of the red

triangles is formed by the three points (0, 0), (1, 1), and (2, 4).

To fi nd the length of the base of one of the red triangles, use

the Distance Formula with the points (0, 0) and (2, 4).

√——

(x2 − x1)2 + (y2 − y1)

2 = √——

( 0 − 2 ) 2 + ( 0 − 4 ) 2

= √—

4 + 16

= √—

20

= 2 √—

5

To fi nd the height, fi rst fi nd the equation of the line

perpendicular to the base through (1, 1). The equation of the

line that contains the base has a slope of 0 − 4

— 0 − 2

= 2 and a

y-intercept of 0, so the equation is y = 2x. Then the slope

of the line perpendicular to the base has a slope of − 1 —

2 and

passes through (1, 1).

y − y1 = m(x − x1)

y − 1 = − 1 —

2 (x − 1)

y − 1 = − 1 —

2 x +

1 —

2

y = − 1 —

2 x +

3 —

2

Next, fi nd the point where the height intersects the base by

solving the system.

y = 2x 2x = − 1 —

2 x +

3 —

2

y = − 1 —

2 x +

3 —

2

⇒

5 —

2 x =

3 —

2

x = 3 —

5

y = 2 ( 3 — 5 ) =

6 —

5

The height of the triangle is the distance between (1, 1) and

( 3 — 5 ,

6 —

5 ) .

√——

(x2 − x1)2 + (y2 − y1)

2 = √——

( 3 — 5 − 1 ) 2 + ( 6 —

5 − 1 ) 2

= √— 4 —

25 +

1 —

25

= √—

5 —

25

= √

— 5 —

5

So, the area of the two red triangles is

2A = 2 ⋅ 1 —

2 bh = 2 ⋅

1 —

2 ⋅ 2 √

— 5 ⋅

√—

5 —

5 = 2.

The common ratio of the geometric series is r = 2 —

8 =

1 —

4 .

The infi nite geometric series that gives the area of the region

has a1 = 8 and r = 1 — 4 , so the series is ∑

i = 1

∞ 8 ( 1 —

4 ) i − 1

.


Chapter 8

29. For the fi rst series, a1 = 20 and r = 10

— 20

= 1 —

2 . So, the sum is

S = a1 —

1 − r =

20 —

1 − 1 — 2

= 40 feet.

For the second series, a1 = 1 and r = 0.5

— 1 = 0.5. So, the sum

is S = a1 —

1 − r =

1 —

1 − 0.5 = 2 seconds.

So, the person will travel 40 feet in 2 seconds.

Also, the tortoise will travel 20 feet in 2 seconds, and with a

20-foot head start it will travel 40 feet in 2 seconds.

Therefore, the person will catch up to the tortoise.

30. Yes, your friend is correct.

0.999 . . . = 0.9 + 0.09 + 0.009 + . . .

In this series, a1 = 0.9 and r = 0.09

— 0.9

= 0.1.

So, the sum of the series is S = a1 —

1 − r =

0.9 —

1 − 0.1 = 1.

31. a. Because side lengths of the removed triangle are half the

lengths of the larger triangle, the area is 1 —

4 the larger area. So,

a1 = 1 —

4 (1) =

1 —

4 , a2 = 3 ⋅

1 —

4 ( 1 —

4 ) =

3 —

16 , a3 = 9 ⋅

1 —

4 ( 1 —

16 ) =

9 —

64 , . . . .

Then, an = 1 —

4 ( 3 —

4 ) n − 1

.

b. ∑ n = 1

∞ an = ∑

n = 1

∞

1 —

4 ( 3 —

4 ) n − 1

=

1 —

4

— 1 −

3 —

4

= 1 ft2

As n increases, the area of the removed triangles gets

closer to the area of the original triangle.


32. x –3 –2 –1 0 1

y 0.5 1.5 4.5 13.5 40.5

× 3 ×3 ×3 ×3

As x increases by 1, y is multiplied by 3. So, the common

ratio is 3, and the data in the table represent an exponential

function.

33. x 0 4 8 12 16

y –7 –1 2 2 –1

6 3 0 –3

–3 –3 –3

The second differences are constant. So, the table represents

a quadratic function.

34. a2 − a1 = −1 − (−7) = 6

a3 − a2 = 5 − (−1) = 6

a4 − a3 = 11 − 5 = 6

a5 − a4 = 17 − 11 = 6

Each difference is 6, so the sequence is arithmetic.

35. a2 − a1 = −1 − 0 = −1

a3 − a2 = −3 − (−1) = −2

The sequence is neither arithmetic nor geometric.

36. a2 — a1

= 40.5

— 13.5

= 3

a3 — a2

= 121.5

— 40.5

= 3

a4 — a3

= 364.5

— 121.5

= 3

Each ratio is 3, so the sequence is geometric.


1. a. 1

A B

2345678

nth Term1n

72 103456

13161922

The sequence is arithmetic because the common

difference is 3.

b. 1

A B

2345678

nth Term1n

52 33456

1-1-3-5


difference is −2.

c. 1

A B

2345678

nth Term1n

12 23456

481632

The sequence is geometric because the common ratio is 2.

d. 1

A B

2345678

nth Term1n

12 1/23456

1/81/1281/32,768

1/2,147,483,648

The sequence is neither geometric nor arithmetic because

there is no common ratio or common difference.


Chapter 8

e. 1

A B

2345678

nth Term1n

32 43456

5678


difference is 1.

f. 1

A B

2345678

nth Term1n

42 13456

-1/2-5/4-13/8-29/16

The sequence is neither arithmetic nor geometric because

there is no common difference or common ratio.

g. 1

A B

2345678

nth Term1n

42 23456

11/21/41/8

The sequence is geometric because the common ratio is 1 —

2 .

h. 1

A B

2345678

nth Term1n

42 53456

9142337

The sequence is neither arithmetic nor geometric, because

there is no common difference or common ratio.

2. a. The fi rst term is 3 and each term is three more than

the previous term. So, a rule for the sequence is

a1 = 3, an = an − 1 + 3.

b. The fi rst term is 18 and each term is four less than


a1 = 18, an = an − 1 − 4.

c. The fi rst term is 3 and each term is twice the previous

term. So, a rule for the sequence is a1 = 3, an = 2an − 1.

d. The fi rst term is 128 and each term is one-half the

previous term. So, a rule for the sequence is

a1 = 128, an = an − 1

— 2 .

e. The fi rst term is 5 and each term is equal to the previous

term. So, a rule for the sequence is a1 = 5, an = an − 1.

f. The fi rst and second terms are and each term is the sum of

the previous two terms. So, a rule for the sequence is

a1 = 1, a2 = 1, an = an−2 + an − 1.

3. a. The fi rst term is 0.5 and each term is twice the previous

term. So, a rule for the sequence is a1 = 0.5, an = 2an − 1.

b. The fi rst term is 1 and each term is 1.5 more than


a1 = 1, an = an − 1 + 1.5.

4. Defi ne the fi rst term(s) of the sequence, then write a rule for

an using the previous term(s).

5. Sample answer: a1 = 2, an = 3an − 1.

The fi rst six terms of the sequence are 2, 6, 18, 54, 162,

and 486.

n

400

an500

300

200

100

04 5 63210

The sequence is geometric.


1. a1 = 3

a2 = a1 − 7 = 3 − 7 = −4

a3 = a2 − 7 = −4 − 7 = −11

a4 = a3 − 7 = −11 − 7 = −18

a5 = a4 − 7 = −18 − 7 = −25

a6 = a5 − 7 = −25 − 7 = −32

2. a0 = 162

a1 = 0.5 ⋅ a0 = 0.5(162) = 81

a2 = 0.5 ⋅ a1 = 0.5(81) = 40.5

a3 = 0.5 ⋅ a2 = 0.5(40.5) = 20.25

a4 = 0.5 ⋅ a3 = 0.5(20.25) = 10.125

a5 = 0.5 ⋅ a4 = 0.5(10.125) = 5.0625

3. f (0) = 1

f (1) = f (0) + 1 = 1 + 1 = 2

f (2) = f (1) + 2 = 2 + 2 = 4

f (3) = f (2) + 3 = 4 + 3 = 7

f (4) = f (3) + 4 = 7 + 4 = 11

f (5) = f (4) + 5 = 11 + 5 =16

4. a1 = 4

a2 = 2a1 − 1 = 2(4) − 1 = 7

a3 = 2a2 − 1 = 2(7) − 1 = 13

a4 = 2a3 − 1 = 2(13) − 1 = 25

a5 = 2a4 − 1 = 2(25) − 1 = 49

a6 = 2a5 − 1 = 2(49) − 1 = 97


Chapter 8

5. n 1 2 3 4 5

an 2 14 98 686 4802

×7 ×7 ×7 ×7

The sequence is geometric with fi rst term a1 = 2 and

common ratio r = 7.

an = r ⋅ an − 1

= 7an − 1

A recursive rule for the sequence is a1 = 2, an = 7an − 1.

6. n 1 2 3 4 5

an 19 13 7 1 −5

−6 −6 −6−6

The sequence is arithmetic with fi rst term a1 = 19 and

common difference d = −6.

an = an − 1 + d

= an − 1 − 6

A recursive rule for the sequence is a1 = 19, an = an − 1 − 6.

7. n 1 2 3 4 5

an 11 22 33 44 55

+11 +11 +11 +11


common difference d = 11.

an = an − 1 + d

= an − 1 + 11

A recursive rule for the sequence is a1 = 11, an = an − 1 + 11.

8. The terms have neither a common difference nor a common

ratio. Beginning with the third term in the sequence, each

term is the product of the previous two terms.

A recursive rule for the sequence is a1 = 1, a2 = 2,

an = an − 1 ⋅ an−2.

9. The fi rst term of the sequence is a1 = 17 − 4(1) = 13 and

the common difference is d = −4.

So, an = an − 1 + d

= an − 1 + (−4)

= an − 1 − 4.


10. The fi rst term of the sequence is a1 = 16(3)1 − 1 = 16 and the

common ratio is r = 3.

So, an = r ⋅ an − 1

= 3an − 1.


11. The recursive rule represents an arithmetic sequence with

fi rst term a1 = −12 and common difference d = 16.

an = a1 + (n − 1)d

= −12 + (n − 1)(16)

= 16n − 28

An explicit rule for the sequence is an = 16n − 28.

12. The recursive rule represents a geometric sequence with fi rst

term a1 = 2 and common ratio r = −6.

an = a1 r n − 1

= 2(−6)n − 1

An explicit rule for the sequence is an = 2(−6)n − 1.

13. If 75% of the fi sh remain each year, then a recursive rule is

a1 = 5200, an = (0.75)an − 1 + 400.

Use a graphing calculator to fi nd that the population of fi sh

stabilizes at about 1600 fi sh.

14. Understand the Problem You are given the conditions

of a loan. You are asked to fi nd the balance after the third

payment and the amount of the last payment.

Make a Plan Because the balance after each payment

depends on the balance after the previous payment, write

a recursive rule that gives the balance after each payment.

Then use a spreadsheet to fi nd the balance after each

payment, rounded to the nearest cent.

Solve the Problem Because the monthly interest rate

is 0.075

— 12

= 0.00625, the balance increases by a factor of

1.00625 each month, and then the payment of $1048.82 is

subtracted.

So, an = (1.00625)an − 1 − 1048.82

Use a spreadsheet and the recursive rule to fi nd the balance

after the third payment and after the 359th payment.

358359

360361

3109.702080.321044.50

357358359

1A B

234

Payment number Balance after payment149,888.68149,776.66149,663.95

123

The balance after the third payment is $149,663.95.

The balance after the 359th payment is $1044.67.

Look Back By continuing the spreadsheet for the 360th

payment, the balance is 2.38.

361360

2.38360

So, it is reasonable that the last payment is $2.38.


Chapter 8



1. A recursive equation tells how the nth term of a sequence is

related to one or more preceding terms.

2. An explicit rule gives an as a function of n and a recursive

rule gives the beginning term(s) of a sequence and a

recursive equation tells how an is related to one or more

preceding terms.


3. a1 = 1

a2 = a1 + 3 = 1 + 3 = 4

a3 = a2 + 3 = 4 + 3 = 7

a4 = a3 + 3 = 7 + 3 = 10

a5 = a4 + 3 = 10 + 3 = 13

a6 = a5 + 3 = 13 + 3 = 16

4. a1 = 1

a2 = a1 − 5 = 1 − 5 = −4

a3 = a2 − 5 = −4 − 5 = −9

a4 = a3 − 5 = −9 − 5 = −14

a5 = a4 − 5 = −14 − 5 = −19

a6 = a5 − 5 = −19 − 5 = −24

5. f (0) = 4

f (1) = 2f (0) = 2(4) = 8

f (2) = 2f (1) = 2(8) = 16

f (3) = 2f (2) = 2(16) = 32

f (4) = 2f (3) = 2(32) = 64

f (5) = 2f (4) = 2(64) = 128

6. f (0) = 10

f (1) = 1 —

2 f (0) =

1 —

2 (10) = 5

f (2) = 1 —

2 f (1) =

1 —

2 (5) = 2.5

f (3) = 1 —

2 f (2) =

1 —

2 (2.5) = 1.25

f (4) = 1 —

2 f (3) =

1 —

2 (1.25) = 0.625

f (5) = 1 —

2 f (4) = 1 — 2 (0.625) = 0.3125

7. a1 = 2

a2 = (a1)2 + 1 = 22 + 1 = 5

a3 = (a2)2 + 1 = 52 + 1 = 26

a4 = (a3)2 + 1 = 262 + 1 = 677

a5 = (a4)2 + 1 = 6772 + 1 = 458,330

a6 = (a5)2 + 1 = 458,3302 + 1 = 210,066,388,901

8. a1 = 1

a2 = (a1)2 − 10 = 12 − 10 = −9

a3 = (a2)2 − 10 = (−9)2 − 10 = 71

a4 = (a3)2 − 10 = 712 − 10 = 5031

a5 = (a4)2 − 10 = 50312 − 10 = 25,310,951

a6 = (a5)2 − 10 = 25,310,9512 − 10 = 640,644,240,524,000

9. f (0) = 2

f (1) = 4

f (2) = f (1) − f (0) = 4 − 2 = 2

f (3) = f (2) − f (1) = 2 − 4 = −2

f (4) = f (3) − f (2) = −2 − 2 = −4

f (5) = f (4) − f (3) = −4 − (−2) = −2

10. f (1) = 2

f (2) = 3

f (3) = f (2) ⋅ f (1) = 3 ⋅ 2 = 6

f (4) = f (3) ⋅ f (2) = 6 ⋅ 3 = 18

f (5) = f (4) ⋅ f (3) = 18 ⋅ 6 = 108

f (6) = f (5) ⋅ f (4) = 108 ⋅ 18 = 1944

11. n 1 2 3 4 5

an 21 14 7 0 −7

−7 −7 −7 −7



an = an − 1 + d

= an − 1 − 7


12. n 1 2 3 4 5

an 54 43 32 21 10

−11 −11 −11 −11



an = an − 1 + d = an − 1 − 11


13. n 1 2 3 4 5

an 3 12 48 192 768

×4 ×4 ×4 ×4


common ratio r = 4.

an = r ⋅ an − 1

= 4an − 1



Chapter 8

14. n 1 2 3 4

an 4 −12 36 −108

×(−3) ×(−3) ×(−3)


common ratio r = −3.

an = r ⋅ an − 1

= (−3) an − 1

A recursive rule for the sequence is a1 = 4, an = (−3) an − 1.

15. n 1 2 3 4 5

an 44 11 11

— 4

11 —

16

11 —

64

× 1 —

4 ×

1 —

4 ×

1 —

4 ×

1 —

4


common ratio r = 1 —

4 .

an = r ⋅ an − 1

= ( 1 — 4 ) an − 1

A recursive rule for the sequence is a1 = 44, an = an−1

— 4 .

16. n 1 2 3 4 5

an 1 8 15 22 29

+7 +7 +7 +7



an = an − 1 + d

= an − 1 + 7



ratio. Beginning with the third term, each term is the product

of the previous two terms. A recursive rule for the sequence

is a1 = 2, a2 = 5, an = an − 2 ⋅ an − 1.


ratio. Beginning with the third term, each term is the product

of the previous two terms.


an = an − 2 ⋅ an − 1.


ratio. Beginning with the third term, each term is the sum of

the previous two terms.


an = an − 2 + an − 1.


ratio. Beginning with the third term, each term is the

difference of the previous two terms.


an = an − 2 − an − 1.


ratio. Note that a1 = 6, a2 = 2 ⋅ 6 = 2 ⋅ a1, a3 = 3 ⋅ 12 =

3 ⋅ a2, a4 = 4 ⋅ 36 = 4 ⋅ a3, and so on.

A recursive rule for the sequence is a1 = 6, an = n ⋅ an − 1.


ratio. Note that a1 = −3, a2 = −3 + 2 = a1 + 2,

a3 = −1 + 3 = a2 + 3, and so on.

A recursive rule for the sequence is a1 = −3, an = an − 1 + n.

23. Make a table of values.

n 1 2 3 4

f (n) 1 2 3 4

+1 +1 +1

The sequence is arithmetic with fi rst term f (1) = 1 and


f (n) = f (n − 1) + d

= f (n − 1) + 1

A recursive rule for the sequence is f (1) = 1, f (n) = f (n – 1) + 1.


n 1 2 3 4

f (n) 8 4 2 1

× 1 — 2 ×

1 — 2 ×

1 — 2

The sequence is geometric with fi rst term f (1) = 8 and


2 .

f (n) = r ⋅ f (n − 1)

= 1 —

2 f (n − 1)

A recursive rule for the sequence is f (1) = 8, f (n) = f (n − 1)

— 2 .


n 1 2 3 4

f (n) −2 1 4 7

+3 +3 +3

The sequence is arithmetic with fi rst term f (1) = −2 and


f (n) = f (n − 1) + d

= f (n – 1) + 3

A recursive rule for the sequence is f (1) = −2,

f (n) = f (n − 1) + 3.


Chapter 8


n 1 2 3 4

f(n) 4 2 0 −2

−2 −2 −2

The sequence is arithmetic with fi rst term f (1) = 4 and


f (n) = f (n − 1) + d

= f (n − 1) − 2

A recursive rule for the sequence is f (1) = 4,

f (n) = f (n − 1) − 2.

27. A recursive rule needs to include the values of the fi rst terms.

So, a recursive rule is a1 = 5, a2 = 2, an = an − 2 − an − 1.

28. The rule does not work for all of the terms. A correct rule is

a1 = 5, a2 = 2, an = an − 2 − an − 1.

29. The explicit rule represents an arithmetic sequence with fi rst

term a1 = 3 + 4(1) = 7 and common difference d = 4.

an = an − 1 + d

= an − 1 + 4



term a1 = −2 − 8(1) = −10 and common difference d = −8.

an = an − 1 + d

= an − 1 − 8

A recursive rule for the sequence is a1 = −10, an = an − 1 − 8.


term a1 = 12 − 10(1) = 2 and common difference d = −10.

an = an − 1 + d

= an − 1 − 10



term a1 = 9 − 5(1) = 4 and common difference d = −5.

an = an − 1 + d

= an − 1 − 5


33. The explicit rule represents a geometric sequence with fi rst

term a1 = 12(11)1−1 = 12 and common ratio r = 11.

an = r ⋅ an − 1

= 11an − 1

A recursive rule for the sequence is a1 = 12, an = 11 an − 1.


term a1 = −7(6)1−1 = −7 and common ratio r = 6.

an = r ⋅ an − 1

= 6an − 1

A recursive rule for the sequence is a1 = −7, an = 6an − 1.


term a1 = 2.5 − 0.6(1) = 1.9 and common difference

d = −0.6.

an = an − 1 + d

= an − 1 − 0.6

A recursive rule for the sequence is a1 = 1.9, an = an − 1 − 0.6.


term a1 = −1.4 + 0.5(1) = −0.9 and common difference

d = 0.5.

an = an − 1 + d

= an − 1 + 0.5

A recursive rule for the sequence is a1 = −0.9,

an = an − 1 + 0.5.


term a1 = − 1 — 2 ( 1 —

4 ) 1−1

= − 1 — 2 and common ratio r =

1 —

4 .

an = r ⋅ an − 1

= 1 —

4 an − 1

A recursive rule for the sequence is a1 = − 1 —

2 , an =

1 —

4 an − 1.


term a1 = 1 —

4 (5)1−1 =

1 —

4 and common ratio r = 5.

an = r ⋅ an − 1

= 5an − 1

A recursive rule for the sequence is a1 = 1 —

4 , an = 5an − 1.


term a1 = 30(1) + 82 = 112 and common difference d = 30.

an = an − 1 + d

= an − 1 + 30

A recursive rule for the amount you have saved n months

from now is a1 = 112, an = an − 1 + 30.

40. The explicit rule represents a geometric sequence with fi rst term

a1 = 35,000 (1.04)1−1 = 35,000 and common ratio r = 1.04.

an = r ⋅ an − 1

= (1.04) an − 1

A recursive rule for your salary is a1 = 35,000, an = 1.04an − 1.


fi rst term a1 = 3 and common difference d = −6.

an = a1 + (n − 1)d

= 3 + (n − 1)(–6)

= 9 − 6n

An explicit rule for the sequence is an = −6n + 9.


fi rst term a1 = 16 and common difference d = 7.

an = a1 + (n − 1)d

= 16 + (n − 1)7

= 7n + 9

An explicit rule for the sequence is an = 7n + 9.


Chapter 8


term a1 = −2 and common ratio r = 3.

an = a1 r n−1

= (−2)(3)n−1

An explicit rule for the sequence is an = −2(3)n−1.


term a1 = 13 and common ratio r = 4.

an = a1 r n − 1

= 13(4) n − 1

An explicit rule for the sequence is an = 13(4)n − 1.


fi rst term a1 = −12 and common difference d = 9.1.

an = a1 + (n − 1)d

= −12 + (n − 1)(9.1)

= 9.1n − 21.1

An explicit rule for the sequence is an = 9.1n − 21.1.


term a1 = −4 and common ratio r = 0.65.

an = a1 r n − 1

= −4 (0.65)n − 1

An explicit rule for the sequence is an = −4(0.65)n − 1.


fi rst term a1 = 5 and common difference d = − 1 — 3 .

an = a1+ (n − 1)d

= 5 + (n − 1) ( − 1 — 3 )

= − 1 — 3 n +

16 —

3

An explicit rule for the sequence is an = − 1 — 3 n +

16 —

3 .


term a1 = −5 and common ratio r = 1 —

4 .

an = a1 r n − 1

= (−5) ( 1 — 4 )

n − 1

An explicit rule for the sequence is an = −5 ( 1 — 4 )

n − 1

.


fi rst term a1 = 20 and common difference d = −2.

an = a1 + (n − 1)(d)

= 20 + (n − 1)(−2)

= −2n + 22

An explicit rule for the number of cans in row n is

an = −2n + 22.


term a1 = 25,600 and common ratio r = 0.86.

an = a1 r n − 1

= 25,600 (0.86) n − 1

An explicit rule for the value of the car after n years is

an = 25,600 (0.86)n − 1.


fi rst term a1 = 4 and common difference d = 6.

an = a1 + (n − 1)d = 4 + (n − 1)(6) = 6n − 2.

An explicit rule is an = 6n − 2.

So, the 1000th term is a1000 = 6(1000) − 2 = 5998.

The answer is B.


term a1 = 0.01 and common ratio r = 1.01.

an = a1 r n − 1 = 0.01 (1.01)n − 1

An explicit rule is an = 0.01 (1.01)n − 1

So, the 873rd term is a873 = 0.01 (1.01)873 − 1 ≈ 58.65

The answer is A.

53. a. Write a recursive rule. Because the company loses 20%

of its members each year, 80% of its members remain in

the company from one year to the next. Also, the company

gains 5000 new members each year.

an = 0.8 ⋅ an − 1 + 5000, where an is the number of

members at the start of the nth year.

A recursive rule is a1 = 50,000, an = 0.8 an − 1 + 5000.

b. Enter 50,000 in a graphing calculator. Then enter the rule

0.8 × Ans + 5000. Press the enter button four times to

fi nd a5 = 35,240.

There are 35,240 members at the start of the fi fth year.

c. Continue pressing enter on the calculator. Observe that the

number of members of the company approaches 25,000.

Over time, the number of members stabilizes at about

25,000 people.

54. a. Write a recursive rule. The initial value is 34. Because

40% of the chlorine evaporates every week, 60% chlorine

remains from one week to the next. Also, 16 ounces of

chlorine is added every week.

an = 0.6an − 1 + 16, where an is the amount of chlorine in

the pool at the start of the nth week.

A recursive rule is a1 = 34, an = 0.6an − 1 + 16.

b. Enter 34 in a graphing calculator. Then enter the rule

0.6 × Ans + 16. Press the enter button twice to fi nd a3 =

37.84. There are 37.84 ounces of chlorine in the pool at

the start of the third week.

c. Continue pressing enter on the calculator. Observe that the

amount of chlorine approaches 40. Overtime, the amount

of chlorine in the pool stabilizes at 40 ounces.

55. Sample answer: You have saved $100 for a vacation. Each

week you save $5 more.

A recursive rule is a1 = 100, an = an − 1 + 5.

56. Sample answer: a1 = 1, a2 = 2, a3 = 3,

an = an − 1 + an − 2 + an − 3.

The fi rst eight terms are 1, 2, 3, 6, 11, 20, 37, and 68.


Chapter 8


of a loan. You are asked to fi nd the balance after the fi fth



depends on the balance after the previous payment, write

a recursive rule that gives the balance after each payment.

Then use a spreadsheet to fi nd the balance after each

payment, rounded to the nearest cent.


is 0.09

— 12

= 0.0075, the balance increases by a factor of

1.0075 each month, and then the payment of $91.37 is

subtracted.

an = 1.0075 an − 1 − 91.37

2223

2425

270.06180.7290.71

212223

1A B

234

Payment number Balance after payment1923.631846.691769.17

123

5 1691.0746 1612.385

a. The balance after the fi fth payment is $1612.38.

b. The balance after the 23rd payment is $90.71, so the

fi nal payment is 1.0075 ($90.71) = $91.39.


of a loan. You are asked to fi nd the balance after the fourth



depends on the balance after the previous payment, write a

recursive rule that gives the balance after each payment. Then

use a spreadsheet to fi nd the balance after each payment.


is 0.115

— 12

, the balance increases by a factor of 1 + 0.115

— 12

each

month, and then the payment of $173.86 is subtracted.

a1 = 10,000, an = ( 1 + 0.115

— 12

) an − 1 − 173.86

8384

343.41172.84

8283

1A B

234

Payment number Balance after payment9921.979843.209763.67

123

5 9683.384

a. The balance after the fourth payment is $9683.38.

b. The balance after the 83rd payment is $172.84, so the

fi nal payment is about

( 1 + 0.115

— 12

) (172.84) = $174.50.

59. Use the recursive formula to create a table of values.

Month 1 2 3 4 5 6 7 8 9 10 11 12

Pairs 1 1 2 3 5 8 13 21 34 55 89 144

When n = 12, f12 = 1 —

√—

5 ( 1 + √

— 5 —

2 ) 12

− 1 —

√—

5 ( 1 − √

— 5 —

2 ) 12

= 144.

In each formula, there are 144 rabbits after one year.

60. a. The initial value is 54,000. Because 2% of the books are

lost or discarded every year, 98% books remain from one

year to the next. Also, the library purchases 1150 new

books each year.

an = 0.98an − 1 + 1150, where an is the number of books

in the library at the beginning of the nth year.

A recursive rule is a1 = 54,000, an = 0.98 an − 1 + 1150.

b.

n

55,000

an

55,500

56,000

56,500

54,500

54,000

53,500

040 50 603020100

The graph appears to approach an = 57,500. So, the

number of books in the library stabilizes at 57,500 books.

61. a. The initial value is 9000. Because 10% of the trees are

harvested each year, 90% of the trees remain from one

year to the next. Also, 800 seedlings are planted each year.

an = 0.9an − 1 + 800, where an is the number of trees on

the farm at the beginning of the nth year.

A recursive rule is a1 = 9000, an = 0.9 an − 1 + 800.

b. After an extended period of time, the number of trees

stabilizes at 8000 trees.

62. a. The initial value is 325. Because 60% of the drug is

removed from the bloodstream every 8 hours, 40% of the

drug remains. Also, 325 milligrams of the drug enters the

bloodstream every 8 hours.

an = 0.4an − 1 + 325, where an is the amount of the drug

in the bloodstream after n doses.

A recursive rule is a1 = 325, an = 0.4an − 1 + 325.

b. After an extended period of time, the amount of the drug

stabilizes at 541 2 —

3 mg. So, the maintenance level of the

drug is 541 2 —

3 milligrams.

c. The maintenance level doubles when the dosage doubles.

The new level is 1083 1 —

3 milligrams.


Chapter 8

63. a. The number of new branches in each stage is 1, 2, 4, 8,

16, 32, and 64.

These numbers form a geometric sequence.

b. The fi rst term is a1 = 1 and common ratio is r = 2. An

explicit rule for the sequence is

an = a1 r n − 1

= 2n − 1.

A recursive rule for the sequence is

a1 = 1, an = 2an − 1.

64. n 1 2 3 4 5 6 7 8 9 10

an 34 17 52 26 13 40 20 10 5 16

n 11 12 13 14 15 16 17 18 19 20

an 8 4 2 1 4 2 1 4 2 1

The terms 4, 2, and 1 eventually repeat.

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14

an 25 76 38 19 58 29 88 44 22 11 34 17 52 26

n 15 16 17 18 19 20 21 22 23 24 25 26 27

an 13 40 20 10 5 16 8 4 2 1 4 2 1

The terms 4, 2, and 1 eventually repeat.

65. Because the monthly interest rate is 0.1

— 12

, the balance

increases by a factor of 1 + 0.1

— 12

each month, and then

the payment of $213.59 is subtracted. Also, the starting

amount is $3500 − $500 = $3000. A recursive rule that

gives the balance after each payment is

a1 = 3000, an = ( 1 + 0.1

— 12

) an − 1 − 213.59.

1415

1617

421.90211.830.01

131415

1A B

23

Payment number Balance after payment2811.412621.25

12

Using the spreadsheet, it takes 15 months to pay back the

loan. The balance after the 14th payment is $211.83, so the

last payment is ( 1 + 0.1

— 12

) (211.83) = $213.60.

66. a. As n increases, the values in the sequence alternate

between positive and negative numbers and get closer

to zero.

b. The set of possible values for r is −1 < r < 0.

The sign of an alternates between positive and negative

and the absolute value decreases.

67. a. The fi rst fi ve terms are

f (1) = 3

f (2) = 10

f (3) = 4 + 2 f (2) − f (1) = 4 + 2(10) − 3 = 21

f (4) = 4 + 2 f (3) − f (2) = 4 + 2(21) − 10 = 36

f (5) = 4 + 2 f (4) − f (3) = 4 + 2(36) − 21 = 55.

b. 3 10 21 36 55

7 11 15 19

4 4 4

The terms of the sequence show a quadratic relationship.

c. Using the quadratic regression feature of a calculator, an

explicit rule for the sequence is an = 2n2 + n.


Sample answer: The Fibonacci sequence is neither arithmetic

nor geometric, but is defi ned by a recursive rule.

69. a. The sequence Tn is 1, 3, 6, 10, . . .. In the nth triangle, the

number of points in each row goes from 1 at the top to

n at the bottom. So, the total number of points in the nth

triangle is ∑ i = 1

n

i = n(n + 1)

— 2 =

1 —

2 n2 +

1 —

2 n.

An explicit rule for Tn is Tn = 1 —

2 n2 +

1 —

2 n.

The sequence Sn is 1, 4, 9, 16, . . ..

In the nth square, there are n rows with n points in each

row. So, an explicit rule for Sn is Sn = n2.

b. The fi rst triangle has 1 point, so T1 = 1. To form the nth

triangle from the previous triangle, add a row of n points.

So, a recursive rule for Tn is T1 = 1, Tn = Tn − 1 + n.

The fi rst square has 1 point, so S1 = 1. To form the nth square

from the previous square, add another row of n − 1 points

and another column of n points. So, a recursive rule for Sn is

Sn = 1, Sn = Sn − 1 + 2n − 1.

c. A rule for the square numbers in terms of the triangular

numbers is Sn = Tn + Tn − 1. To see this, separate the nth

square above the diagonal as shown.

2 3 4

The points in the lower left portion form the nth

triangle, and the points in the upper right portion form

the (n − 1)th triangle. So, the total number of points is

Tn + Tn − 1.


Chapter 8

70. a. Because you expect to earn an annual return of 8%

on your savings, the balance increases by a factor of

1.08 each year, and then $30,000 is subtracted.

A recursive equation is an = 1.08an − 1 − 30,000.

b. an = 1.08an − 1 − 30,000

an + 30,000 = 1.08 an − 1

an + 30,000 —

1.08 = an − 1

Use a spreadsheet to fi nd a0.

1A B

234567891011121314

16

ann

27,777.78020

53,497.9419

77,312.9118

99,363.8117

119,781.311615141312111098

15 76

138,686.40156,191.11172,399.18187,406.65201,302.45214,168.94226,082.35237,113.29247,327.12256,784.3717 5265,541.0818 4273,649.1519 3281,156.6220 2288,107.9821 1

22 0 294,544.4323

So, a0 is about $294,544.


71. √— x + 2 = 7 Check √

— 25 + 2 =

? 7

√—

x = 5 5 + 2 =? 7

( √—

x ) 2 = 52 7 = 7 ✓

x = 25

72. 2 √—

x − 5 = 15 Check 2 √—

100 − 5 =? 15

2 √—

x = 20 2 ⋅ 10 − 5 =? 15

√—

x = 10 15 = 15 ✓

( √—

x ) 2 = 102

x = 100

73. 3 √—

x + 16 = 19 Check 3 √—

27 + 16 =?

19

3 √—

x = 3 3 + 16 =?

19

( 3 √—

x ) 3= 33 19 = 19 ✓

x = 27

74. 2 3 √—

x − 13 = −5 Check 2 3 √—

64 − 13 =?

−5

2 3 √—

x = 8 2 ⋅ 4 − 13 =?

−5

3 √—

x = 4 −5 = −5 ✓

( 3 √—

x ) 3 = 43

x = 64

75. y = a —

x

9 = a —

2

18 = a

The inverse variation equation is y = 18

— x .

When x = 4, y = 18

— 4 =

9 —

2 .

76. y = a —

x

3 = a —

−4

−12 = a

The inverse variation equation is y = −12

— x .

When x = 4, y = −12

— 4 = −3.

77. y = a —

x

32 = a —

10

320 = a

The inverse variation equation is y = 320

— x .

When x = 4, y = 320

— 4 = 80.

8.4−8.5 What Did You Learn? (p. 451)

1. Label the horizontal axis as n and the vertical axis as Sn to

show n is the independent variable.

2. The number 0.999 . . . can be rewritten as

0.9 + 0.09 + 0.009 + . . . , which is an infi nite geometric

series with a1 = 0.9 and r = 0.1.

Because ∣ r ∣ < 1, the series has the sum S = 0.9 —

1 − 0.1 =

0.9 —

0.9 = 1.

3. The starting salary is a1 and the raise is r − 1 written as a

percent.

4. The recursive rule does not make sense because when n = 5,

harvesting 8656.1 trees does not make sense.

Chapter 8 Review (pp. 452−454)

1. The pattern is a1 = 1 ⋅ 2, a2 = 2 ⋅ 3, a3 = 3 ⋅ 4,

a4 = 4 ⋅ 5, . . ..

So, a rule for the nth layer is an = n 2 + n.

2. A pattern for the terms is 3(1) + 4, 3(2) + 4, 3(3) + 4, . . . ,

3(12) + 4.

So, 7 + 10 + 13 + . . . + 40 = ∑ i=1

12

(3i + 4) .

3. A pattern for the terms is 02 + 0, 12 + 1, 22 + 2, 32 + 3, . . ..

So, 0 + 2 + 6 + 12 + . . . = ∑ i=0

∞ (i 2 + i) .


Chapter 8

4. ∑ i=2

7

(9 − i 3) = (9 − 23) + (9 − 33) + (9 − 43) +

(9 − 53) + (9 − 63) + (9 − 73)

= 1 + (−18) + (−55) + (−116) + (−207) + (−334)

= −729

5. ∑ i=1

46

i = 1 + 2 + 3 + . . . + 46

= 46(46 + 1)

— 2

= 1081

6. ∑ i=1

12

i 2 = 12 + 22 + . . . + 122

= 12(12 + 1)(2 ⋅ 12 + 1)

—— 6

= 650

7. ∑ i=1

5

3 + i

— 2 = ( 3 + 1

— 2 ) + ( 3 + 2

— 2 ) + ( 3 + 3

— 2 ) +

( 3 + 4 —

2 ) + ( 3 + 5

— 2 )

= 2 + 2.5 + 3 + 3.5 + 4

= 15

8. 12 4 −4 −12 −20

−8 −8 −8 −8

Because the terms have a common difference of −8, the

sequence is arithmetic.

9. The terms of the sequence are 6(1) − 4, 6(2) − 4, 6(3) − 4,

6(4) − 4, . . . .

So, a rule for the nth term is an = 6n − 4.

n

an

10

5

0

15

25

20

30

4 5 6 7310 2

10. an = a1 + (n − 1)d

a14 = a1 + (14 − 1)(d)

42 = a1 + 13(3)

3 = a1

A rule for the nth term is an = a1 + (n − 1)d

= 3 + (n − 1)3

= 3n.

n

an

10

5

0

15

25

20

4 5 6310 2

11. Use the terms to write a linear system.

a6 = a1 + (6 − 1)d −12 = a1 + 5d

a12 = a1 + (12 − 1)d ⇒ −36 = a1 + 11d

−12 = a1 + 5d

−36 = a1 + 11d

24 = −6d

−4 = d

−12 = a1 + 5(−4)

8 = a1

A rule for the nth term is an = a1 + (n − 1)d

= 8 + (n − 1)(−4)

= −4n + 12.

n

an

10

5

0

−5

−10

5 631 2

12. The fi rst term is a1 = 2 + 3(1) = 5.

The last term is a36 = 2 + 3(36) = 110.

So, the sum of the series is 2070. Find the sum.

S36 = 36 ( 5 + 110 —

2 ) = 2070

13. Let an be the salary in the nth year.

a1 = 37,000 and d = 1500. So, an = a1 + (n − 1)d

= 37,000 + (n − 1)(1500)

= 1500n + 35,500.

∑ n=1

6

an = ∑ n=1

6

(1500n + 35,500)

= S6

= 6 ( a1 + a6 —

2 )

= 6 [ 1500(1) + 35,500 + 1500(6) + 35,500 ————

2 ]

= 244,500

Your total earnings in 6 years is $244,500.

14. 7 14 28 56 112

×2 ×2 ×2 ×2

The terms have a common ratio of 2, so the sequence is

geometric.


Chapter 8

15. a1 = 25, r = 10

— 25

= 2 —

5

an = a1r n−1

= 25 ( 2 — 5 )

n−1

A rule for the nth term is an = 25 ( 2 — 5 )

n−1

.

n

an

10

5

0

15

25

20

4 5 6310 2

16. an = a1r n−1

a5 = a1r 5−1

162 = a1 (−3)4

2 = a1

A rule for the nth term is an = a1r n−1 = 2(−3)n−1.

n

100

0

−100

−200

−300

−400

−500

5 631 2

17. a3 = a1r 3−1 16 = a1r 2

a5 = a1r 5−1 ⇒

256 = a1r 4

16 = a1r 2 ⇒ 16

— r 2

= a1

256 = a1r 4 ⇒ 256 = 16

— r 2

r 4

256 = 16r 2

16 = r 2

±4 = r

16 —

(±4)2 = a1

1 = a1

There are two possible rules for the nth term. One rule is

an = a1r n−1 = 4 n−1, and the other is an = (−4) n−1.

an = 4 n−1 an = (−4) n−1

n

an

400

200

0

600

1000

800

4 5 6310 2

n

an

−600

−900

−300

62

18. ∑ i=1

9

5 (−2)i−1 = 5(−2)0 + 5(−2)1 + 5(−2)2 + 5(−2)3 + . . .

+ 5(−2)8

= 5 − 10 + 20 − 40 + 80 − 160 + 320 −

640 + 1280

= 855

19. The partial sums are

S1 = 1

S2 = 1 + ( − 1 — 4 ) = 0.75

S3 = 1 + ( − 1 — 4 ) +

1 —

16 ≈ 0.81

S4 = 1 + ( − 1 — 4 ) +

1 —

16 + ( −

1 — 64 ) ≈ 0.80

S5 = 1 + ( − 1 — 4 ) +

1 —

16 + ( −

1 — 64 ) +

1 —

256 ≈ 0.80.

As n increases, Sn approaches 0.80

n

Sn

0.4

0.2

0

0.6

1.0

0.8

4 5 6310 2

20. For this series, a1 = −2 and r = 1 —

2

— −2

= − 1 —

4 .

Because ∣ − 1 —

4 ∣ < 1, the sum of the series exists.


S = a1 —

1−r =

−2 —

1 − ( − 1 —

4 )

= −1.6.

21. 0.1212 … = 0.12 + 0.0012 + . . .

= 12

— 100

+ 12 —

10,000 + . . .

For this series, a1 = 12

— 100

and r = 1 —

100 .

The sum of the series is S = a1 —

1−r =

12

— 100

— 1 −

1 —

100

= 12

— 99

= 4 —

33 . . .

22. a1 = 7 23. a1 = 6

a2 = 7 + 11 = 18 a2 = 4 ⋅ 6 = 24

a3 = 18 + 11 = 29 a3 = 4 ⋅ 24 = 96

a4 = 29 + 11 = 40 a4 = 4 ⋅ 96 = 384

a5 = 40 + 11 = 51 a5 = 4 ⋅ 384 = 1536

a6 = 51 + 11 = 62 a6 = 4 ⋅ 1536 = 6144


Chapter 8

24. f (0) = 4

f (1) = 4 + 2 ⋅ 1 = 6

f (2) = 6 + 2 ⋅ 2 = 10

f (3) = 10 + 2 ⋅ 3 = 16

f (4) = 16 + 2 ⋅ 4 = 24

f (5) = 24 + 2 ⋅ 5 = 34

25. 9 6 4 8 —

3

16 —

9

× 2 —

3 ×

2 —

3 ×

2 —

3 ×

2 —

3

The sequence is geometric with a1 = 9 and common ratio

r = 2 —

3 .

an = r ⋅ an − 1

= 2 —

3 an − 1

A recursive rule for the sequence is a1 = 9, an = 2 —

3 an−1.


ratio.

Note that a1 = 2, a2 = 2 = 2(2 − 1) = a1 (2 − 1),

a3 = 4 = 2(3 − 1) = a2 (3 − 1), . . . .

A recursive rule for the sequence is a1 = 2, an = an−1 (n − 1).


ratio. Beginning with the third term, each term is the

difference of the previous two terms.


an = an−2 − an −1.

28. The sequence is geometric with fi rst term

a1 = 105 ( 3 — 5 )

1−1

= 105 and common ratio r = 3 —

5 .

an = r ⋅ an−1

= 3 —

5 an−1

A recursive rule for the sequence is a1 = 105, an = 3 —

5 an−1.



an = a1 + (n − 1)d

= −4 + (n − 1)(26)

= 26n − 30

An explicit rule for the sequence is an = 26n − 30.


common ratio r = −5.

an = a1r n−1

= 8(−5)n−1

An explicit rule for the sequence is an = 8(−5) n−1.



5 .

an = a1r n−1

= 26 ( 2 — 5 )

n−1

An explicit rule for the sequence is an = 26 ( 2 — 5 )

n−1

.

32. Because the population increases about 4% each year, it

increases by a factor of 1.04 each year.

A recursive rule is P1 = 11,120, Pn = 1.04Pn−1.

33. The fi rst hexagon has 1 dot. To form the nth hexagon from

the previous hexagon, add four sides to the hexagon. Use

n dots to form the fi rst side, and n − 1 dots to form each

of the remaining sides. The total number of dots added is

n + 3(n − 1) = 4n − 3.

So, a recursive rule for the nth hexagonal number is

a1 = 1, an = an − 1 + 4n − 3.

Chapter 8 Test (p. 455)

1. The fi rst term is a1 = 6(1) − 13 = −7.

The last term is a24 = 6(24) − 13 = 131.

So, the sum is S24 = 24 ( −7 + 131 —

2 ) = 1488.

2. ∑ n = 1

16

n2 = 16(16+1)(2 ⋅ 16+1)

—— 6

= 1496

3. ∑ k = 1

∞ 2(0.8) k−1 =

2 —

1 − 0.8

= 2 —

0.2

= 10

4. ∑ i = 1

6

4(−3) i − 1 = 4 + 4(−3) + 4(−3)2 + 4(−3)3 + 4(−3)4

+ 4(−3)5

= 4 − 12 + 36 − 108 + 324 − 972

= −728

5. The graph represents a geometric sequence because the

points appear to follow an exponential curve. The terms are

2 1−1, 2 2−1, 2 3−1, and 2 4−1. So, a rule for the nth term is

an = 2 n−1.

6. The graph represents an arithmetic sequence because the

points lie on a line. The terms are 13 − 2(1), 13 − 2(2),

13 − 2(3), and 13 − 2(4). So, a rule for the nth term is

an = −2n + 13.


Chapter 8

7. The graph is neither an arithmetic sequence nor a geometric

sequence because the points do not lie on a line nor follow

an exponential curve.

The terms are a1 = 1 —

4 , a2 =

5 —

12 =

1 —

4 +

1 —

6 = a1 +

1 —

3 ⋅ 2 ,

a3 = 1 —

2 =

5 —

12 +

1 —

12 = a2 +

1 —

3 ⋅ 2 2 and a4 =

13 —

24 =

1 —

2 +

1 —

24 =

a3 + 1 —

3 ⋅ 23 . So, a recursive rule for the nth term is

an = an−1 + 1 —

3 ⋅ 2n−1 .

8. an = r ⋅ an−1

= 1 —

2 ⋅ an−1

A recursive rule is a1 = 32, an = 1 —

2 an−1. Use a table to fi nd a9.

n 1 2 3 4 5 6 7 8 9

an 32 16 8 4 2 1 1 —

2

1 —

4

1 —

8

So, a9 = 1 —

8 .

9. a1 = 2 + 7(1) = 9

The sequence is arithmetic with d = 7.

an = an−1 + d

= an − 1 + 7

So, a recursive rule is a1 = 9, an = an − 1 + 7.

Use a table to fi nd a9.

n 1 2 3 4 5 6 7 8 9

an 9 16 23 30 37 44 51 58 65

So, a9 = 65.

10. The terms are a1 = 2, a2 = 0 = a1 − 2, a3 = −3 = a2 − 3,

a4 = −7 = a3 −4, a5 = −12 = a4 − 5, . . . .

So, a recursive rule is a1 = 2, an = an − 1 − n.

n 1 2 3 4 5 6 7 8 9

an 2 0 −3 −7 −12 −18 −25 −33 −42

So, a9 = −42.

11. 5 −20 80 −320 1280

×(−4) ×(−4) ×(−4) ×(−4)

an = r ⋅ an − 1

= −4an − 1

A recursive rule is a1 = 5, an = −4an − 1.

The sequence is geometric with a1 = 5 and r = −4.

an = a1r n−1

= 5(−4) n−1

So, an explicit rule is an = 5(−4) n−1.

12. Yes, b is half of the sum of a and c.

Sample answer: Let d be the common difference of the

sequence. Then b = a + d and c = b + d = a + 2d.

So, 1 —

2 (a + c) =

1 —

2 (a + a + 2d ) =

1 —

2 (2a + 2d) = a + d = b.

13. a. n represents the number of rows and the number of

columns.

an represents the number of blue squares.

b. n an

1 1

2 2

3 5

4 8

5 13

6 18

7 25

8 32

c. n an = n2 —

2 + 1 —

4 [ 1 − (−1)n ]

1 1

2 2

3 5

4 8

5 13

6 18

7 25

8 32

The values in this table are the same as the values in the table

in part (b).

14. This situation represents a series because it is represented by

the sum of values.

15. The length of each loop of the spring is given by 16, 16(0.9),

16(0.9)2, . . . .

So, the total length is ∑ n = 1

∞ 16(0.9)n−1 .

The series is geometric with common ratio r = 0.9.

Because ∣ r ∣ < 1, the series will have a fi nite sum.

So, the length of the spring is fi nite.

∑ n = 1

∞ 16(0.9) n−1 = 16

— 1 − 0.9

= 160

So, the length of the spring is 160 inches.


Chapter 8

Chapter 8 Standards Assessment (pp. 456−457)

1. a8 = 392 and a10 = 440.

Because the sequence is geometric, an = a1r n−1.

a8 = a1r 8−1 392 = a1r 7

a10 = a1r 10−1 ⇒

440 = a1r 9

392 = a1r 7

392 —

r 7 = a1

440 = 392

— r 7

r 9

440 = 392r 2

55 —

49 = r 2

± √

— 55 —

7 = r

Because the frequencies are positive, use the positive root

to fi nd a1.

392 = a1 ( √—

55 — 7 ) 7

a1 ≈ 261.64

So, an = 261.64 ( √—

55 —

7 ) n−1

≈ 261.64(1.06)n−1.

When n = 4, a4 ≈ 311.

The answer is B.

2. a. Because the interest rate is 0.0075 per month, the balance

increases by a factor of 1.0075 each month, and then

the payment of $300 is subtracted. A recursive rule is

a1 = 16,000, an = 1.0075 an−1 − 300.

b. Use a spreadsheet to fi nd the balance at the start of the

18th month.

1819

12,952.1912,749.33

1718

1A B

234

Month Balance16,00015,820

15,638.65

123

At the beginning of the 18th month you owe $12,749.33.

c. Continue the spreadsheet until the balance is less than zero.

6970

71

405.98109.02

6869

-190.1670

The balance is less than zero at the start of the 70th month.

So, the last payment is 300 − 190.16 = $109.84 in the

69th month. It will take 69 months to pay off the loan.

d. The balance at the beginning of each month is

an = 1.0075 an−1 − 350. Use a spreadsheet to fi nd the

number of payments.

5859

66.26-283.24

5758

1A B

234

Month Balance16,000.0015,770.0015,538.28

123

The balance is less than zero at the start of the 58th

month. So, the last payment is 350 − 283.24 = $66.76 in

the 57th month. It will take 57 months to pay off the loan.

When you pay $350 a month, you pay a total of

350(56) + 66.76 = $19,666.76.

When you pay $300 a month, you pay a total of

300(68) + 109.84 = $20,509.84.

So, you save 20,509.84 − 19,666.76 = $843.08

3. The products F ⋅ l are equal to 1500 for each pair of values.

So, an equation is F = 1500

— l .

F and l are related by inverse variation.

4. For A, the average rate of change is

f (4) − f (1)

— 4 − 1

= 4 √

— 6 − 4 √

— 3 —

3

≈ 0.96.

For B, the function is y = 10

— x . The average rate of change is

10

— 4 −

10 —

1

— 4 − 1 =

−30

— 4

— 3

= −2.5.

For C, the average rate of change is 11 − 2

— 4 − 1

= 3.

For D, the average rate of change is 5 − (−4)

— 4 − 1

= 3.

The order is B, A, C, and D or B, A, D, and C.

5. a. The sequence is arithmetic, because the common

difference is 1.22 meters.

b. The fi rst term is a1 = 36.5 and d = 1.22.

an = a1 + (n − 1)d

= 36.5 + (n − 1)(1.22)

= 1.22n + 35.28

A rule for the sequence is an = 1.22n + 35.28.

c. The curve radius in the outside lane is

a8 = 1.22(8) + 35.28

= 45.04 meters.

Yes, the track shown meets the requirement.


Chapter 8

6.

FunctionPositive

real zerosNegativereal zeros

Imaginaryzeros

Totalzeros

f 3 0 2 5

g 3 1 0 4

h 5 1 2 8

For f, the number of sign changes is 3 or 5.

For g, the number of sign changes is 3.

For h, the number of sign changes is 5 or 7.

The number of sign changes is at least the number of positive

real zeroes according to DeCarte’s Rule of Signs, or more by

an even number.

7. The total downward distance traveled by the basketball is

∑ n = 1

∞ 10(0.36) n − 1 . The total upward distance is

∑ n = 1

∞ 3.6(0.36) n − 1 . So, the total distance traveled by the

baseball is

∑ n = 1

∞ 10(0.36) n −1 + ∑

n = 1

∞ 3.6(0.36) n −1 =

10 —

1 − 0.36 +

3.6 —

1 − 0.36

≈ 15.63 + 5.63

= 21.26 feet.

The total downward distance traveled by the baseball is

∑ n = 1

∞ 10(0.3) n − 1 . The total upward distance is ∑

n = 1

∞ 3(0.3) n − 1 .

So, the total distance traveled by the baseball is

∑ n = 1

∞ 10(0.3) n −1 + ∑

n = 1

∞ 3(0.3) n −1 =

10 —

1 − 0.3 +

3 —

1 − 0.3

≈ 14.29 + 4.29

= 18.58 feet.

The difference is about 21.26 − 18.58 = 2.68 feet.

The answer is C.

8. a. x + √—

−16 = 0

x = − √—

−16

x = −4i

The solution is a pure imaginary number.

b. (11 − 2i) − (−3i + 6) = 8 + x

11 − 2i + 3i − 6 = 8 + x

5 + i = 8 + x

−3 + i = x

The solution is an imaginary number.

c. 3x 2 − 14 = −20

3x 2 = −6

x 2 = −2

x = ± √—

−2

x = ± √—

2 i

The solutions are pure imaginary numbers.

d. x 2 + 2x = −3

x 2 + 2x + 1 = −2

(x + 1)2 = −2

x + 1 = ± √—

−2

x = −1 ± √—

2 i

The solutions are imaginary numbers.

e. x 2 = 16

x = ± √—

16

x = ±4

The solutions are real numbers.

f. x 2 − 5x − 8 = 0

Use the Quadratic Formula.

x = −(−5) ± √——

(−5)2 −4(1)(−8) ———

2(1)

= 5 ± √

— 25 + 32 ——

2

= 5 —

2 ±

√—

57 —

2

The solutions are real numbers.

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