aat solutions - ch12
DESCRIPTION
mathematicsTRANSCRIPT
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667Algebra 2
Worked-Out Solution Key
Prerequisite Skills (p. 792)
1. The domain of f (x) is all real numbers except x Þ 0.
2. The range of g(x) is all real numbers.
3. The composition f (g(x)) is equal to 1 }
4x 1 2 .
4. 7x 1 3 5 31 5. 9 5 2x 2 7
7x 5 28 16 5 2x
x 5 4 8 5 x
6. 14 5 23x 1 8 7. 10 2 3x 5 28
6 5 23x 23x 5 18
22 5 x x 5 26
8. 11x 1 9 5 3x 1 17 9. 2x 1 3 5 26 2 x
8x 5 8 3x 5 29
x 5 1 x 5 23
10. 3x 1 y 5 0 3 4 12x 1 4y 5 0
22x 2 4y 5 230 22x 2 4y 5 230
10x 5 230
x 5 23
3(23) 1 y 5 0 → y 5 9
The solution is (23, 9).
11. 2x 2 2y 5 10 2x 2 2y 5 10
x 1 y 5 210 3 2 2x 1 2y 5 220
4x 5 210
x 5 2 5 } 2
2 5 } 2 1 y 5 210 → y 5 2
15 } 2
The solution is 1 2 5 } 2 , 2
15 } 2 2 .
12. 4x 2 5y 5 25 4x 2 5y 5 25
0.5x 1 1.5y 5 18.5 3 28 24x 2 12y 5 2148
217y 5 2123
y 5 123
} 17
4x 2 5 1 123 }
17 2 5 25 → x 5
260 } 17
The solution is 1 260 }
17 ,
123 }
17 2 .
13. f (g(x)) 5 2(22x21) 2 1 5 24x21 2 1
Domain: all real numbers except x 5 0
14. f ( f (x)) 5 2(2x 2 1) 2 1 5 4x 2 3
Domain: all real numbers
15. g(g(x)) 5 22(22x21)21 5 x
Domain: all real numbers except x 5 0
Lesson 12.1
12.1 Guided Practice (pp. 794–797)
1. a1 5 1 1 4 5 5 2. f (1) 5 (22)1 2 1 5 1
a2 5 2 1 4 5 6 f (2) 5 (22)2 2 1 5 22
a3 5 3 1 4 5 7 f (3) 5 (22)3 2 1 5 4
a4 5 4 1 4 5 8 f (4) 5 (22)4 2 1 5 28
a5 5 5 1 4 5 9 f (5) 5 (22)5 2 1 5 16
a6 5 6 1 4 5 10 f (6) 5 (22)6 2 1 5 232
3. a1 5 1 } 1 1 1 5
1 } 2
a2 5 2 } 2 1 1 5
2 } 3
a3 5 3 } 3 1 1 5
3 } 4
a4 5 4 } 4 1 1 5
4 } 5
a5 5 5 } 5 1 1 5
5 } 6
a6 5 6 } 6 1 1 5
6 } 7
4. The sequence 3, 8, 15, 24, . . .
5
n21
an
can be written as 1 p 3, 2 p 4, 3 p 5, 4 p 6, . . .. The next term is a5 5 5 p 7 5 35. A rule for the nth term is an 5 n(n 1 2).
5. a9 5 92 5 81
There are 81 apples in the 9th layer.
6. ai 5 5i
Lower limit 5 1
Upper limit 5 20
Summation notation: i 5 1
∑ 20
5i
7. ai 5 i2
} i2 1 1
Lower limit 5 1
Upper limit 5 infi nity
Summation notation: i 5 1
∑ `
i2 }
i2 1 1
8. ai 5 6i
Lower limit 5 1
Upper limit 5 infi nity
Summation notation: i 5 0
∑ `
6i
9. 4 1 i
Lower limit 5 1
Upper limit 5 8
Summation notation: i 5 1
∑ 8
(4 1 i)
Chapter 12
n2ws-1200-a.indd 667 6/27/06 11:30:51 AM
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668Algebra 2Worked-Out Solution Key
10. i 5 1
∑ 5
8i 5 8(1) 1 8(2) 1 8(3) 1 8(4) 1 8(5)
5 8 1 16 1 24 1 32 1 40
5 120
11. k 5 3
∑ 7
(k2 2 1) 5 (32 2 1) 1 (42 2 1) 1 (52 2 1) 1
(62 2 1) 1 (72 2 1) 5 8 1 15 1 24 1 35 1 48
5 130
12. i 5 1
∑ 34
1 5 34
13. n 5 1
∑ 6
n 5 1 1 2 1 3 1 4 1 5 1 6 5 21
14. n 5 1
∑ 9
n2 5 12 1 22 1 32 1 42 1 52 1 62 1 72 1 82 1 92
5 1 1 4 1 9 1 16 1 25 1 36 1 49 1 64 1 81
5 285
There are 285 apples in the stack.
12.1 Exercises (pp. 798–800)
Skill Practice
1. Another name for summation notation is sigma notation.
2. A sequence is a list of numbers and a series is the sum of the terms in a sequence.
3. a1 5 1 1 2 5 3 4. a1 5 6 2 1 5 5
a2 5 2 1 2 5 4 a2 5 6 2 2 5 4
a3 5 3 1 2 5 5 a3 5 6 2 3 5 3
a4 5 4 1 2 5 6 a4 5 6 2 4 5 2
a5 5 5 1 2 5 7 a5 5 6 2 5 5 1
a6 5 6 1 2 5 8 a6 5 6 2 6 5 0
5. a1 5 12 5 1 6. f (1) 5 13 1 2 5 3
a2 5 22 5 4 f (2) 5 23 1 2 5 10
a3 5 32 5 9 f (3) 5 33 1 2 5 29
a4 5 42 5 16 f (4) 5 43 1 2 5 66
a5 5 52 5 25 f (5) 5 53 1 2 5 127
a6 5 62 5 36 f (6) 5 63 1 2 5 218
7. a1 5 41 2 1 5 1 8. a1 5 212 5 21
a2 5 42 2 1 5 4 a2 5 222 5 24
a3 5 43 2 1 5 16 a3 5 232 5 29
a4 5 44 2 1 5 64 a4 5 242 5 216
a5 5 45 2 1 5 216 a5 5 252 5 225
a6 5 46 2 1 5 1024 a6 5 262 5 236
9. f (1) 5 12 2 5 5 24 10. a1 5 (1 1 3)2 5 16
f (2) 5 22 2 5 5 21 a2 5 (2 1 3)2 5 25
f (3) 5 32 2 5 5 4 a3 5 (3 1 3)2 5 36
f (4) 5 42 2 5 5 11 a4 5 (4 1 3)2 5 49
f (5) 5 52 2 5 5 20 a5 5 (5 1 3)2 5 64
f (6) 5 62 2 5 5 31 a6 5 (6 1 3)2 5 81
11. f (1) 5 2 4 } 1 5 24 12. a1 5
3 } 1 5 3
f (2) 5 2 4 } 2 5 22 a2 5
3 } 2
f (3) 5 2 4 } 3 a3 5
3 } 3 5 1
f (4) 5 2 4 } 4 5 21 a4 5
3 } 4
f (5) 5 2 4 } 5 a5 5
3 } 5
f (6) 5 2 4 } 6 5 2
2 } 3 a6 5
3 } 6 5
1 } 2
13. a1 5 2(1)
} 1 1 2 5 2 } 3 14. f (1) 5
1 } 2(1) 2 1 5 1
a2 5 2(2)
} 2 1 2 5 1 f (2) 5 2 } 2(2) 2 1 5
2 } 3
a3 5 2(3)
} 3 1 2 5 6 } 5 f (3) 5
3 } 2(3) 2 1 5
3 } 5
a4 5 2(4)
} 4 1 2 5 4 } 3 f (4) 5
4 } 2(4) 2 1 5
4 } 7
a5 5 2(5)
} 5 1 2 5 10
} 7 f (5) 5 5 } 2(5) 2 1 5
5 } 9
a6 5 2(6)
} 6 1 2 5 3 } 2 f (6) 5
6 } 2(6) 2 1 5
6 } 11
15. Given terms: 5 p 1 2 4, 5 p 2 2 4, 5 p 3 2 4, 5 p 4 2 4, . . .
Next term: 5 p 5 2 4 5 21
Rule for nth term: 5n 2 4
16. Given terms: 21 2 1, 22 2 1, 23 2 1, 24 2 1, . . .
Next term: 25 2 1 5 16
Rule for nth term: 2n 2 1
17. Given terms: (21)1(4 p 1), (21)2(4 p 2), (21)3(4 p 3), (1)4(4 p 4), . . .
Next term: (21)5(4 p 5) 5 220
Rule for nth term: (21)n(4n)
18. Given terms: 13 1 1, 23 1 1, 33 1 1, 43 1 1, . . .
Next term: 53 1 1 5 126
Rule for nth term: n3 1 1
19. Given terms: 2 }
3 p 1 , 2 }
3 p 2 , 2 }
3 p 3 , 2 }
3 p 4 , . . .
Next term: 2 }
3 p 5 5 2 } 15
Rule for nth term: 2 }
3n
20. Given terms: 1 p 2
} 1 1 2
, 2 p 2
} 2 1 2
, 3 p 2
} 3 1 2
, 4 p 2
} 4 1 2
, . . .
Next term: 5 p 2
} 5 1 2
5 10
} 7
Rule for nth term: 2n }
2 1 n
21. Given terms: 1 }
4 ,
2 }
4 ,
3 }
4 ,
4 }
4 ,
5 }
4 , . . .
Next term: 6 }
4
Rule for nth term: n }
4
Chapter 12, continued
n2ws-1200-a.indd 668 6/28/06 1:48:37 PM
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669Algebra 2
Worked-Out Solution Key
Chapter 12, continued
22. Given terms: 2 p 1 2 1
} 1 p 10
, 2 p 2 2 1
} 2 p 10
, 2 p 3 2 1
} 3 p 10
, 2 p 4 2 1
} 4 p 10
, . . .
Next term: 2 p 5 2 1
} 5 p 10
5 9 } 50
Rule for nth term: 2n 2 1
} 10n
23. Given terms: 2.4 1 0.7(1), 2.4 1 0.7(2), 2.4 1 0.7(3), 2.4 1 0.7(4), . . .
Next term: 2.4 1 0.7(5) 5 5.9
Rule for nth term: 2.4 1 0.7n
24. Given terms: 5.8 2 1.6(1), 5.8 2 1.6(2), 5.8 2 1.6(3), 5.8 2 1.6(4), 5.8 2 1.6(5), . . .
Next term: 5.8 2 1.6(6) 5 23.8
Rule for nth term: 5.8 2 1.6n
25. Given terms: 0.2 1 12, 0.2 1 22, 0.2 1 32, 0.2 1 42, . . .
Next term: 0.2 1 52 5 25.2
Rule for nth term: 0.2 1 n2
26. Given terms: 1.2 1 7.8(1), 1.2 1 7.8(2), 1.2 1 7.8(3), 1.2 1 7.8(4), . . .
Next term: 1.2 1 7.8(5) 5 40.2
Rule for nth term: 1.2 1 7.8n
27. D;
a1 5 1(1 1 1)
} 2 5 1
a2 5 2(2 1 1)
} 2 5 3
a3 5 3(3 1 1)
} 2 5 6
a4 5 4(4 1 1)
} 2 5 10
28.2
n21
an 29.
n1
7
an
30.
3
n21
an
31. an
4
n21
32.
4
n21
an 33.
4
n21
an
34.
5
n21
an 35.
1
n21
an
36.
1
n21
an
37. ai 5 3i 1 4
Lower limit 5 1
Upper limit 5 5
Summation notation: i 5 1
∑ 5
(3i 1 4)
38. 6i 2 1
Lower limit 5 1
Upper limit 5 5
Summation notation: i 5 1
∑ 5
(6i 2 1)
39. 2i 2 3
Lower limit 5 1
Upper limit 5 infi nity
Summation notation: i 5 1
∑ `
(2i 2 3)
n2ws-1200-a.indd 669 6/28/06 1:49:13 PM
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670Algebra 2Worked-Out Solution Key
40. (22)i
Lower limit 5 1
Upper limit 5 infi nity
Summation notation: i 5 1
∑ `
(22)i
41. 7i 2 4
Lower limit 5 1
Upper limit 5 infi nity
Summation notation: i 5 1
∑ `
(7i 2 4)
42. 1 }
3i
Lower limit 5 1
Upper limit 5 4
Summation notation: i 5 1
∑ 4
1 }
3i
43. i }
3 1 i
Lower limit 5 1
Upper limit 5 7
Summation notation: i 5 1
∑ 7
i }
3 1 i
44. i2 2 2
Lower limit 5 1
Upper limit 5 infi nity
Summation notation: i 5 1
∑ `
(i2 2 2)
45. i 5 1
∑ 6
2i 5 2 p 1 1 2 p 2 1 2 p 3 1 2 p 4 1 2 p 5 1 2 p 6
5 2 1 4 1 6 1 8 1 10 1 12
5 42
46. i 5 1
∑ 5
7i 5 7 p 1 1 7 p 2 1 7 p 3 1 7 p 4 1 7 p 5
5 7 1 14 1 21 1 28 1 35
5 105
47. n 5 0
∑ 4
n3 5 03 1 13 1 23 1 33 1 43
5 0 1 1 1 8 1 27 1 64
5 100
48. k 5 1
∑ 4
3k2 5 3 p 12 1 3 p 22 1 3 p 32 1 3 p 42
5 3 1 12 1 27 1 48
5 90
49. k 5 3
∑ 6
(5k 2 2) 5 (5 p 3 2 2) 1 (5 p 4 2 2)
1 (5 p 5 2 2) 1 (5 p 6 2 2)
5 13 1 18 1 23 1 28
5 82
50. n 5 1
∑ 5
(n2 2 1) 5 (12 2 1) 1 (22 2 1) 1 (32 2 1) 1 (42 2 1) 1 (52 2 1) 5 0 1 3 1 8 1 15 1 24
5 50
51. i 5 1
∑ 8
2 } i 5
2 } 1 1
2 } 2 1
2 } 3 1
2 } 4 1
2 } 5 1
2 } 6 1
2 } 7 1
2 } 8
5 2 1 1 1 2 } 3 1
1 } 3 1
1 } 2 1
1 } 4 1
2 } 5 1
2 } 7
5 761
} 140
52. k 5 1
∑ 6
k }
k 1 1 5
1 } 1 1 1 1
2 } 2 1 1 1
3 } 3 1 1
1 4 } 4 1 1 1
5 } 5 1 1 1
6 } 6 1 1
5 1 } 2 1
2 } 3 1
3 } 4 1
4 } 5 1
5 } 6 1
6 } 7
5 617
} 140
53. i 5 1
∑ 35
1 5 35
54. n 5 1
∑ 16
n 5 1 1 2 1 3 1 4 1 5 1 6 1 . . . 1 14 1 15 1 16
5 136
55. i 5 1
∑ 25
i 5 1 1 2 1 3 1 4 1 5 1 . . . 1 23 1 24 1 25
5 325
56. n 5 1
∑ 18
n2 5 12 1 22 1 32 1 42
1 52 1 . . . 1 162 1 172 1 182
5 1 1 4 1 9 1 16
1 25 1 . . . 1 256 1 289 1 324
5 2109
57. The fi rst term is missing. The lower limit is 0, so the fi rst term should be 2(0) 1 3, or 3. Correct sum: 3 1 5 1 7 1 9 1 11 1 13 5 48
58. B;
i 5 1
∑ 20
i 5 1 1 2 1 3 1 4 1 . . . 1 17 1 18 1 19 1 20
5 210
59. true;
i 5 1
∑ n
kai 5 ka1 1 ka2 1 ka3 1 ka4 1 . . . 1 kan
5 k(a1 1 a2 1 a3 1 . . . 1 an)
5 k i 5 1
∑ n
ai
60. true;
i 5 1
∑ n
(ai 1 bi) 5 (a1 1 b1) 1 (a2 1 b2)
1 (a3 1 b3) 1 . . . 1 (an 1 bn)
5 (a1 1 a2 1 a3 1 . . . 1 an)
1 (b1 1 b2 1 b3 1 . . . 1 bn)
5 i 5 1
∑ n
ai 1 i 5 1
∑ n
bi
Chapter 12, continued
n2ws-1200-a.indd 670 6/27/06 11:31:12 AM
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671Algebra 2
Worked-Out Solution Key
61. false;
Sample answer: Let ai 5 2i and bi 5 24i.
i 5 1
∑ 3
(2i)(24i) 5 (2 p 1)(24 p 1) 1 (2 p 2)(24 p 2)
1 (3 p 2)(24 p 3)
5 28 1 (232) 1 (272)
5 2112
i 5 1
∑ 3
2i 5 2 p 1 1 2 p 2 1 2 p 3 5 12
i 5 1
∑ 3
(24i) 5 24 p 1 1 (24 p 2) 1 (24 p 3) 5 224
1 i 5 1
∑ 3
2i 2 1 i 5 1
∑ 3
(24i) 2 5 12(224) 5 2288
Because 2112 Þ 2288, i 5 1
∑ n
aibi Þ 1 i 5 1
∑ n
ai 2 1 i 5 1
∑ n
bi 2 . 62. false;
Sample answer: Let ai 5 i and k 5 2.
i 5 1
∑ 3
i2 5 12 1 22 1 32 5 1 1 4 1 9 5 14
1 i 5 1
∑ 3
i 2 2
5 (1 1 2 1 3)2 5 62 5 36
Because 14 Þ 36, i 5 1
∑ n
aik Þ 1
i 5 1 ∑
n
ai 2 k.Problem Solving
63. a3 5 180(3 2 2)
} 3 5 608
a4 5 180(4 2 2)
} 4 5 908
a5 5 180(5 2 2)
} 5 5 1088
a6 5 180(6 2 2)
} 6 5 1208
a7 5 180(7 2 2)
} 7 ø 128.578
Tn 5 180(n 2 2)
} n p n 5 180(n 2 2)
Total measure of angles in skylight 5 180(12 2 2), or 18008.
64. Use Special Series Formula for sum of fi rst n positive
integers: i 5 1
∑ n
i 5 n(n 1 1)
} 2
In 100 days, you will have 100(100 1 1)
}} 2 5 5050 pennies,
or $50.50.
To save $500, you need 50,000 pennies.
n(n 1 1)
} 2 5 50,000
n(n 1 1) 5 100,000
n2 1 n 2 100,000 5 0
(n 2 Ï}
100,000 )2 5 0
n 5 Ï}
100,000
5 316.23
You need about 316 days to save $500.
65. Given terms: 21 2 1, 22 2 1, 23 2 1, 24 2 1, 25 2 1
Rule for nth term: an 5 2n 2 1
a6 5 26 2 1 5 63
a7 5 27 2 1 5 127
a8 5 28 2 1 5 255
You need 63 moves to move 6 rings, 127 moves to move 7 rings, and 255 moves to move 8 rings.
66. a. d4 5 0.3(2)4 2 2 1 0.4 5 1.6
Mars is 1.6 astronomical units from the sun.
b. 149,600,000 km
}} 1 a.u.
3 1.6 a.u. 5 239,360,000 km
Mars is about 239,360,000 kilometers from the sun.
c.
Position of planet from sun
Mea
n d
ista
nce
fro
m s
un
(a.
u.)
0 1 2 3 4 5 6 7 8 9 n
dn
048
12162024283236
67. a. a5 5 5(5 1 1)
} 2 5 15
There are 15 balls in the fi fth layer.
b. a1 1 a2 1 a3 1 a4 1 a5
5 1(1 1 1)
} 2 1 2(2 1 1)
} 2 1 3(3 1 1)
} 2 1 4(4 1 1)
} 2 1 5(5 1 1)
} 2
5 1 1 3 1 6 1 10 1 15
5 35
There are 35 balls in the stack.
c. The difference in each layer is
n2 2 n(n 1 1)
} 2 5 2n2 2 n2 2 n
}} 2
5 n2 2 n
} 2
5 n(n 2 1)
} 2 .
Each layer of the square pyramid contains exactly
n(n 2 1)
} 2 more balls than the corresponding layer in the
triangular pyramid.
68. The number of balls in the top n layers is
i 5 1
∑ n
n(n 1 1)
} 2 5
i 5 1 ∑
n
n2 1 n
} 2 .
Using Exercise 59: i 5 1
∑ n
n2 1 n
} 2 5
1 } 2 i 5 1
∑ n
(n2 1 n)
Using Exercise 60: 1 }
2 i 5 1
∑ n
(n2 1 n) 5 1 } 2 1
i 5 1 ∑
n
n2 1 i 5 1
∑ n
n 2 Using the special formulas on page 797:
1 }
2 1
i 5 1 ∑
n
n2 1 i 5 1
∑ n
n 2 5 1 } 2 1 n(n 1 1)(2n 1 1)
}} 6 1
n(n 1 1) } 2 2
Chapter 12, continued
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672Algebra 2Worked-Out Solution Key
Mixed Review
69. 18 5 4x 2 2 70. 23 5 6x 2 1
20 5 4x 24 5 6x
5 5 x 4 5 x
71. 12 5 2 2 5x
10 5 25x
22 5 x
72. 5 5 8x 2 5 73. 17 5 20 2 2x
10 5 8x 23 5 22x
5 }
4 5 x
3 }
2 5 x
74. 14 5 5x 1 4 75. 7 2 3x 5 16
10 5 5x 23x 5 9
2 5 x x 5 23
76. 9 1 8x 5 25 77. 11x 2 6 5 239
8x 5 16 11x 5 233
x 5 2 x 5 23
78. d 5 Ï}}
(6 2 4)2 1 (1 2 (25))2
5 Ï}
4 1 36
5 2 Ï}
10
79. d 5 Ï}}}
(22 2 (27))2 1 (21 2 4)2
5 Ï}
25 1 25
5 5 Ï}
2
80. d 5 Ï}}
(5 2 0)2 1 (22 2 5)2
5 Ï}
25 1 49
5 Ï}
74
81. d 5 Ï}}
(1 2 (24))2 1 (9 2 6)2
5 Ï}
25 1 9
5 Ï}
34
82. d 5 Ï}}}
(6 2 2)2 1 (24 2 (25))2
5 Ï}
16 1 1
5 Ï}
17
83. d 5 Ï}}}
(22 2 (25))2 1 (28 2 (24))2
5 Ï}
9 1 16
5 5
84. d 5 Ï}}
(5 2 9)2 1 (6 2 7)2
5 Ï}
16 1 1
5 Ï}
17
85. d 5 Ï}}
(3 2 (21))2 1 (2 2 8)2
5 Ï}
16 1 36
5 2 Ï}
13
86. d 5 Ï}}
(29 2 4)2 1 (26 2 0)2
5 Ï}
169 1 36
5 Ï}
205
Graphing Calculator Activity 12.1 (p. 801)
1. a. 5, 9, 13, 17, 21, 25, 29, 33, 37, 41
b.
c. Sum 5 230
2. a. 9, 12, 15, 18, 21, 24, 27, 30, 33, 36
b.
c. Sum 5 225
3. a. 32, 29, 26, 23, 20, 17, 14, 11, 8, 5
b.
c. Sum 5 185
4. a. 17, 19, 21, 23, 25, 27, 29, 31, 33, 35
b.
c. Sum 5 260
5. a. 4, 7, 12, 19, 28, 39, 52, 67, 84, 103
b.
c. Sum 5 415
6. a. 1, 2, 4, 8, 16, 32, 64, 128, 256, 512
b.
c. Sum 5 1023
Chapter 12, continued
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673Algebra 2
Worked-Out Solution Key
Lesson 12.2
12.2 Guided Practice (pp. 802–805)
1. a2 2 a1 5 14 2 17 5 23
a3 2 a2 5 11 2 14 5 23
a4 2 a3 5 8 2 11 5 23
a5 2 a4 5 5 2 8 5 23
Each difference is 23, so the sequence is arithmetic.
2. a1 5 17; d 5 14 2 17 5 23
an 5 17 1 (n 2 1)(23) 5 20 2 3n
a20 5 20 2 3(20) 5 240
3. a11 5 a1 1 (11 2 1) d
257 5 a1 1 10(27)
13 5 a1
an 5 13 1 (n 2 1)(27) 5 20 2 7n
a20 5 20 2 7(20) 5 2120
4. a7 5 26 5 a1 1 (7 2 1)d → 26 5 a1 1 6d
a16 5 71 5 a1 1 (16 2 1)d → 71 5 a1 1 15d
45 5 9d
5 5 d
71 5 a1 1 15(5) → a1 5 24
an 5 24 1 (n 2 1)(5) 5 29 1 5n
a20 5 29 1 5(20) 5 91
5. a1 5 2 1 7(1) 5 9
a12 5 2 1 7(12) 5 86
S12 5 12 1 9 1 86 }
2 2 5 570
6. a1 5 3(1) 5 3
a8 5 3(8) 5 24
S8 5 8 1 3 1 24 }
2 2 5 108
The house of cards contains 108 cards.
12.2 Exercises (pp. 806–809)
Skill Practice
1. The constant difference between consecutive terms of an arithmetic sequence is called the common difference.
2. An arithmetic sequence is a list of numbers that have the same common difference between consecutive terms. An arithmetic series is the sum of the terms.
3. a2 2 a1 5 22 2 1 5 23
a3 2 a2 5 25 2 (22) 5 23
a4 2 a3 5 28 2 (25) 5 23
a5 2 a4 5 211 2 (28) 5 23
The sequence is arithmetic with common difference 23.
4. a2 2 a1 5 14 2 16 5 22
a3 2 a2 5 11 2 14 5 23
The differences are not constant, so the sequence is not arithmetic.
5. a2 2 a1 5 14 2 5 5 9
a3 2 a2 5 23 2 14 5 9
a4 2 a3 5 32 2 23 5 9
a5 2 a4 5 41 2 32 5 9
The sequence is arithmetic with common difference 9.
6. a2 2 a1 5 27 2 (210) 5 3
a3 2 a2 5 25 2 (27) 5 2
The differences are not constant, so the sequence is not arithmetic.
7. a2 2 a1 5 1 2 0.5 5 0.5
a3 2 a2 5 1.5 2 1 5 0.5
a4 2 a3 5 2 2 1.5 5 0.5
a5 2 a4 5 2.5 2 2 5 0.5
The sequence is arithmetic with common difference 0.5.
8. a2 2 a1 5 10 2 20 5 210
a3 2 a2 5 5 2 10 5 25
The differences are not constant, so the sequence is not arithmetic.
9. a2 2 a1 5 5 } 4 2
7 } 4 5 2
2 } 4
a3 2 a2 5 3 } 4 2
5 } 4 5 2
2 } 4
a4 2 a3 5 2 3 } 4 2
3 } 4 5 2
6 } 4
The differences are not constant, so the sequence is not arithmetic.
10. a2 2 a1 5 2 } 7 2
1 } 7 5
1 } 7
a3 2 a2 5 4 } 7 2
2 } 7 5
2 } 7
The differences are not constant, so the sequence is not arithmetic.
11. a2 2 a1 5 21 2 1 2 5 } 2 2 5
3 } 2
a3 2 a2 5 1 } 2 2 (21) 5
3 } 2
a4 2 a3 5 2 2 1 } 2 5
3 } 2
a5 2 a4 5 7 } 2 2 2 5
3 } 2
The sequence is arithmetic with common difference 3 }
2 .
12. a1 5 1; d 5 4 2 1 5 3
an 5 1 1 (n 2 1)(3) 5 3n 2 2
a20 5 3(20) 2 2 5 58
13. a1 5 5; d 5 11 2 5 5 6
an 5 5 1 (n 2 1)(6) 5 6n 2 1
a20 5 6(20) 2 1 5 119
14. a1 5 8; d 5 21 2 8 5 13
an 5 8 1 (n 2 1)(13) 5 13n 2 5
a20 5 13(20) 2 5 5 255
Chapter 12, continued
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674Algebra 2Worked-Out Solution Key
15. a1 5 23; d 5 21 2 (23) 5 2
an 5 23 1 (n 2 1)(2) 5 2n 2 5
a20 5 2(20) 2 5 5 35
16. a1 5 6; d 5 2 2 6 5 24
an 5 6 1 (n 2 1)(24) 5 24n 1 10
a20 5 24(20) 1 10 5 270
17. a1 5 25; d 5 14 2 25 5 211
an 5 25 1 (n 2 1)(211) 5 211n 1 36
a20 5 211(20) 1 36 5 2184
18. a1 5 0; d 5 2 } 3 2 0 5
2 } 3
an 5 0 1 (n 2 1) 1 2 } 3 2 5
2 } 3 n 2
2 } 3
a20 5 2 } 3 (20) 2
2 } 3 5
38 } 3
19. a1 5 2; d 5 2 5 } 3 2 2 5 2
1 } 3
an 5 2 1 (n 2 1) 1 2 1 } 3 2 5 2
1 } 3 n 1
7 } 3
a20 5 2 1 } 3 (20) 1
7 } 3 5 2
13 } 3
20. a1 5 1.5; d 5 3.6 2 1.5 5 2.1
an 5 1.5 1 (n 2 1)(2.1) 5 2.1n 2 0.6
a20 5 2.1(20) 2 0.6 5 41.4
21. The general rule for an arithmetic sequence is an 5 a1 1 (n 2 1)d, not an 5 a1 1 nd. So,
an 5 37 1 (n 2 1)(213) 5 213n 1 50.
22. The values of 37 and 213 were substituted in the wrong place. It should be an 5 37 1 (n 2 1)(213), so an 5 213n 1 50.
23. a16 5 a1 1 (16 2 1)d 3
n1
an
52 5 a1 1 15(5)
223 5 a1
an 5 223 1 (n 2 1)(5)
an 5 5n 2 28
n 1 2 3 4 5 6
an 223 218 213 28 23 2
24. a6 5 a1 1 (6 2 1)d 8
n
an
21 216 5 a1 1 5(9)
261 5 a1
an 5 261 1 (n 2 1)(9)
an 5 9n 2 70
n 1 2 3 4 5 6
an 261 252 243 234 225 216
25. a4 5 a1 1 (4 2 1)d
14
n1
an
96 5 a1 1 3(214)
138 5 a1
an 5 138 1 (n 2 1)(214)
an 5 214n 1 152
n 1 2 3 4 5 6
an 138 124 110 96 82 68
26. a12 5 a1 1 (12 2 1)d
8
n21
an
23 5 a1 1 11(27)
74 5 a1
an 5 74 1 (n 2 1)(27)
an 5 27n 1 81
n 1 2 3 4 5 6
an 74 67 60 53 46 39
27. a10 5 a1 1 (10 2 1)d
2
n21
an
30 5 a1 1 9 1 7 } 2 2
2 3 } 2 5 a1
an 5 2 3 } 2 1 (n 2 1) 1 7 }
2 2
an 5 7 } 2 n 2 5
n 1 2 3 4 5 6
an 21.5 2 5.5 9 12.5 16
Chapter 12, continued
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675Algebra 2
Worked-Out Solution Key
28. a11 5 a1 1 (11 2 1)d
1
n21
an
1 }
2 5 a1 1 10 1 2
1 } 2 2
11
} 2 5 a1
an 5 11
} 2 1 (n 2 1) 1 2 1 } 2 2
an 5 6 2 1 } 2 n
n 1 2 3 4 5 6
an 5.5 5 4.5 4 3.5 3
29. C;
a30 5 a1 1 (30 2 1)d
57 5 a1 1 29(4)
259 5 a1
an 5 259 1 (n 2 1)(4)
an 5 263 2 4n
30. a10 5 a1 1 (10 2 1)d → 85 5 a1 1 9d
a4 5 a1 1 (4 2 1)d → 31 5 a1 1 3d
54 5 6d
9 5 d
85 5 a1 1 9(9) → a1 5 4
an 5 4 1 (n 2 1)(9) 5 25 1 9n
31. a14 5 a1 1 (14 2 1)d → 79 5 a1 1 13d
a6 5 a1 1 (6 2 1)d → 39 5 a1 1 5d
40 5 8d
5 5 d
79 5 a1 1 13(5) → a1 5 14
an 5 14 1 (n 2 1)(5) 5 9 1 5n
32. a17 5 a1 1 (17 2 1)d → 40 5 a1 1 16d
a3 5 a1 1 (3 2 1)d → 22 5 a1 1 2d
42 5 14d
3 5 d
40 5 a1 1 16(3) → a1 5 28
an 5 28 1 (n 2 1)(3) 5 211 1 3n
33. a20 5 a1 1 (20 2 1)d → 258 5 a1 1 19d
a8 5 a1 1 (8 2 1)d → 210 5 a1 1 7d
248 5 12d
24 5 d
258 5 a1 1 19(24) → a1 5 18
an 5 18 1 (n 2 1)(24) 5 22 2 4n
34. a15 5 a1 1 (15 2 1)d → 137 5 a1 1 14d
a9 5 a1 1 (9 2 1)d → 89 5 a1 1 8d
48 5 6d
8 5 d
137 5 a1 1 14(8) → a1 5 25
an 5 25 1 (n 2 1)(8) 5 17 1 8n
35. a11 5 a1 1 (11 2 1)d → 35 5 a1 1 10d
a2 5 a1 1 (2 2 1)d → 17 5 a1 1 d
18 5 9d
2 5 d
35 5 a1 1 10(2) → a1 5 15
an 5 15 1 (n 2 1)(2) 5 13 1 2n
36. a12 5 a1 1 (12 2 1)d → 29 5 a1 1 11d
a7 5 a1 1 (7 2 1)d → 4 5 a1 1 6d
213 5 5d
2 13
} 5 5 d
29 5 a1 1 11 1 2 13
} 5 2 → a1 5 98
} 5
an 5 98
} 5 1 (n 2 1) 1 2 13
} 5 2 5 111
} 5 2 13
} 5 n
37. a9 5 a1 1 (9 2 1)d → 24 5 a1 1 8d
a5 5 a1 1 (5 2 1)d → 15 5 a1 1 4d
9 5 4d
9 }
4 5 d
24 5 a1 1 8 1 9 } 4 2 → a1 5 6
an 5 6 1 (n 2 1) 1 9 } 4 2 5
15 } 4 2
9 } 4 n
38. a11 5 a1 1 (11 2 1)d → 22 5 a1 1 10d
a6 5 a1 1 (6 2 1)d → 0 5 a1 1 5d
22 5 5d
2 2 } 5 5 d
22 5 a1 1 10 1 2 2 } 5 2 → a1 5 2
an 5 2 1 (n 2 1) 1 2 2 } 5 2 5
12 } 5 2
2 } 5 n
39. B;
a13 5 a1 1 (13 2 1)d → 248 5 a1 1 12d
a6 5 a1 1 (6 2 1)d → 26 5 a1 1 5d
242 5 7d
26 5 d
248 5 a1 1 12(26) → a1 5 24
an 5 24 1 (n 2 1)(26) 5 30 2 6n
40. a1 5 1 1 3(1) 5 4
a10 5 1 1 3(10) 5 31
S10 5 10 1 4 1 31 }
2 2 5 175
41. a1 5 23 2 2(1) 5 25
a8 5 23 2 2(8) 5 219
S8 5 8 1 25 1 (219) }
2 2 5 296
42. a1 5 14 2 6(1) 5 8
a18 5 14 2 6(18) 5 294
S18 5 18 1 8 1 (294) }
2 2 5 2774
Chapter 12, continued
n2ws-1200-a.indd 675 6/27/06 11:31:46 AM
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676Algebra 2Worked-Out Solution Key
43. a1 5 29 1 11(1) 5 2
a22 5 29 1 11(22) 5 233
S22 5 22 1 2 1 233 }
2 2 5 2585
44. a3 5 72 2 6(3) 5 54
a9 5 72 2 6(9) 5 18
S7 5 7 1 54 1 18 }
2 2 5 252
45. a5 5 254 1 9(5) 5 29
a14 5 254 1 9(14) 5 72
S10 5 10 1 29 1 72 }
2 2 5 315
46. a1 5 2
a15 5 58
S15 5 15 1 2 1 58 }
2 2 5 450
47. a1 5 21
a8 5 34
S8 5 8 1 21 1 34 }
2 2 5 132
48. a1 5 44
a7 5 2
S7 5 7 1 2 1 44 }
2 2 5 161
49. 2, 7, 12, 17, . . .
a1 5 2, d 5 5
an 5 2 1 (n 2 1)(5) 5 23 1 5n
50. 21, 24, 27, 210, . . .
a1 5 21, d 5 23
an 5 21 1 (n 2 1)(23) 5 2 2 3n
51. 23, 25, 27, 29, . . .
a1 5 23, d 5 22
an 5 23 1 (n 2 1)(22) 5 21 2 2n
52. The graph of an is a scatter plot whose points lie on the line that represents the graph of f (x).
53. false; Sample answer: Consider the series 2 1 4 1 6 1 8, whose sum is 20. If d is doubled, the series becomes 2 1 6 1 10 1 14, whose sum is 32. Because 32 Þ 2(20), the statement is false.
54. true;
Because an is an arithmetic sequence, the fi rst three terms a, b, and c, are a, a 1 n, and a 1 2n.
b 0 1 }
2 (a 1 c)
a 1 n 0 1 }
2 (a 1 (a 1 2n))
a 1 n 0 1 }
2 (2a 1 2n)
a 1 n 5 a 1 n ✓
55. i 5 1
∑ n
(25 1 7i) 5 486
n1 a1 1 an }
2 2 5 486
na1 1 nan 5 486(2)
n(2) 1 n(25 1 7n) 5 972
7n2 2 3n 2 972 5 0
(7n 1 81)(n 2 12) 5 0
n 5 2 81
} 7 or n 5 12
The number of terms must be positive, so n 5 12.
56. i 5 1
∑ n
(10 2 3i) 5 228
n1 a1 1 an }
2 2 5 228
na1 1 nan 5 228(2)
n(7) 1 n(10 2 3n) 5 256
23n2 1 17n 1 56 5 0
(23n 2 7)(n 2 8) 5 0
n 5 8 or n 5 2 7 } 3
The number of terms must be positive, so n 5 8.
57. i 5 1
∑ n
(58 2 8i) 5 21150
n1 a1 1 an }
2 2 5 21150
na1 1 nan 5 21150(2)
n(50) 1 n(58 2 8n) 5 22300
28n2 1 108n 1 2300 5 0
24(n 2 25)(2n 1 23) 5 0
n 5 25 or n 5 2 23
} 2
The number of terms must be positive, so n 5 25.
58. i 5 1
∑ n
(5 2 5i) 5 250
n1 a1 1 an }
2 2 5 250
na1 1 nan 5 250(2)
n(0) 1 n(5 2 5n) 5 2100
25n2 1 5n 1 100 5 0
25(n 2 5)(n 1 4) 5 0
n 5 5 or n 5 24
The number of terms must be positive, so n 5 5.
Chapter 12, continued
n2ws-1200-a.indd 676 6/27/06 11:31:50 AM
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677Algebra 2
Worked-Out Solution Key
59. i 5 3
∑ n
(23 2 4i) 5 2507
(n 2 2) 1 a3 1 an }
2 2 5 2507
(n 2 2)(a3 1 an) 5 2507(2)
(n 2 2)[215 1 (23 2 4n)] 5 21014
(n 2 2)(218 2 4n) 5 21014
218n 2 4n2 1 36 1 8n 5 21014
24n2 2 10n 1 1050 5 0
22(n 2 15)(2n 1 35) 5 0
n 5 15 or n 5 217.5
The number of terms must be positive, so n 5 15.
60. i 5 5
∑ n
(7 1 12i) 5 455
(n 2 4) 1 a5 1 an }
2 2 5 455
(n 2 4)(a5 1 an) 5 455(2)
(n 2 4)[67 1 (7 1 12n)] 5 910
(n 2 4)(74 1 12n) 5 910
74n 1 12n2 2 296 2 48n 5 910
12n2 1 26n 2 1206 5 0
2(6n 1 67)(n 2 9) 5 0
n 5 9 or n 5 2 67
} 6
The number of terms must be positive, so n 5 9.
61. i 5 1
∑ 150
(2i 2 1) 5 n 1 a1 1 a150 }
2 2
5 150 1 1 1 299 }
2 2
5 22,500
62. S3 5 n 1 a1 1 a3 }
2 2
3 2 x 1 x 1 1 2 3x 5 3 1 3 2 x 1 1 2 3x }}
2 2
23x 1 4 5 3 1 24x 1 4 } 2 2
23x 1 4 5 26x 1 6
3x 5 2
x 5 2 } 3
a1 5 3 2 2 } 3 5
7 } 3
a2 5 2 } 3
a3 5 1 2 3 1 2 } 3 2 5 21
Because the common difference is 2 5 } 3 , the next term is
21 2 5 } 3 , or 2
8 } 3 .
Problem Solving
63. a. a1 5 6, d 5 6
an 5 6 1 (n 2 1)(6)
an 5 6n
b. Total number of cells 5 S9 1 1
S9 5 9 1 a1 1 a9 }
2 2
S9 5 9 1 6 1 54 }
2 2
S9 5 270
There are 271 cells in the honeycomb.
64. a1 5 3, d 5 2 S7 5 7 1 a1 1 a7 }
2 2
an 5 3 1 (n 2 1)(2) S7 5 7 1 3 1 15 }
2 2
an 5 1 1 2n S7 5 63
There are 1 1 2n band members in each row. The band has 63 members.
65. a. a1 5 4, d 5 8
an 5 4 1 (n 2 1)(8)
an 5 24 1 8n
b. S12 5 12 1 a1 1 a12 }
2 2
S12 5 12 1 4 1 92 }
2 2
S12 5 576
There are 576 visible blocks.
66. a. n d(n)
1 D(1) 2 D(0) 5 16(12) 2 16(02) 5 16
2 D(2) 2 D(1) 5 16(22) 2 16(12) 5 48
3 D(3) 2 D(2) 5 16(32) 2 16(22) 5 80
4 D(4) 2 D(3) 5 16(42) 2 16(32) 5 112
b. a1 5 16, d 5 32
an 5 16(n 2 1)(32) 5 216 1 32n
So, a rule is d(n) 5 216 1 32n.
c.
12
n21
d(n)
Chapter 12, continued
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678Algebra 2Worked-Out Solution Key
Chapter 12, continued
67. S5 5 1000, n 5 5, d 5 50
an 5 a1 1 (n 2 1)d
a5 5 a1 1 (5 2 1)(50)
a5 5 a1 1 200
S5 5 5 1 a1 1 a5 }
2 2
1000 5 5 1 a1 1 a1 1 200 }}
2 2
1000 1 2 } 5 2 5 2a1 1 200
400 2 200 5 2a1
100 5 a1
$100 should be given away on the fi rst day.
68. a. n dn (in.) ln (in.)
1 2 2π
2 2.008 2.008π
3 2.016 2.016π
4 2.024 2.024π
Sample calculation:
d2 5 2 1 0.004(1)(2) 5 2.008
b. l1, l2, l3, . . . is an arithmetic sequence.
ln 5 l1 1 (n 2 1)(d)
ln 5 2π 1 (n 2 1)(0.008π)
ln 5 [2 1 (n 2 1)(0.008)]π
c. a1 5 2.008, d 5 0.008, an 5 5
an 5 a1 1 (n 2 1)d
5 5 2.008 1 (n 2 1)(0.008)
3 5 0.008n
375 5 n
The paper must be wrapped around the dowel 375 times.
ln 5 [2 1 (n 2 1)(0.008)]π (from part b)
l375 5 4.992π
l1 5 2π
S375 5 375 1 2π 1 4.992π }
2 2 ø 4118.6
The length of paper in the roll is about 4118.6 inches.
d. For a roll with a 7-inch diameter:
7 5 2.008 1 (n 2 1)(0.008)
5 5 0.008n
625 5 n
l1 5 2π
l625 5 [2 1 (625 2 1)(0.008)]π 5 6.992π
S625 5 625 1 2π 1 6.992π }
2 2 ø 8827.9
The length of paper in the 7-inch diameter roll is about 8827.9 inches. This is about 2.14 times the length of paper in the 5-inch diameter roll, so you could expect to pay about 2.14($1.50) 5 $3.21 for the 7-inch diameter roll.
Mixed Review
70. x1/5 5 7 71. x2/3 5 36
(x1/5)5 5 75 (x2/3)3/2 5 363/2
x 5 16,807 x 5 216
72. 6x2/5 2 5 5 19
6x2/5 5 24
x2/5 5 4
(x2/5)5/2 5 45/2
x 5 32
73. 3x3/5 2 4 5 5
3x3/5 5 9
x3/5 5 3
(x3/5)5/3 5 35/3
x 5 3 p 32/3
x 5 3 3 Ï}
32
x 5 3 3 Ï}
9
74. (x 1 10)1/4 5 5 75. (x 2 3)3/4 5 64
[(x 1 10)1/4]4 5 54 [(x 2 3)3/4]4/3 5 644/3
x 1 10 5 625 x 2 3 5 256
x 5 615 x 5 259
76. 6x 5 216 77. 5x 5 32
log6 6x 5 log6 216 log5 5
x 5 log5 32
x 5 log6 216 x 5 log5 32
x 5 log 216
} log 6
x 5 log 32
} log 5
x 5 3 x ø 2.153
78. 104x 2 6 5 12 79. 105x 1 1 1 5 5 19
104x 5 18 105x 1 1 5 14
log 104x 5 log 18 log 105x 1 1 5 log 14
4x 5 log 18 5x 1 1 5 log 14
x ø 0.314 x ø 0.029
80. 7x 2 3 5 49x 2 8
7x 2 3 5 (72)x 2 8
7x 2 3 5 72x 2 16
x 2 3 5 2x 2 16
13 5 x
81. 34x 1 1 5 729x
34x 1 1 5 (36)x
34x 1 1 5 36x
4x 1 1 5 6x
1 5 2x
1 }
2 5 x
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679Algebra 2
Worked-Out Solution Key
82. Mean: } x 5 5 1 6 1 6 1 6 1 8 1 9
}} 6 5 40
} 6 5 6 2 }
3
Median: 6
Mode: 6
83. Mean: } x 5 32 1 36 1 38 1 43 1 43 1 45 1 46
}}} 7 5 283
} 7
ø 40.43
Median: 43
Mode: 43
84. Mean: } x 5 75 1 80 1 81 1 82 1 83 1 92 1 92 1 92
}}}} 8
5 677
} 8 5 84.625
Median: 82 1 83
} 2 5 82.5
Mode: 92
85. Mean: } x 5 25 1 (24) 1 (23) 1 (22) 1 (21) 1 1 1 5
}}}} 7
5 29
} 7 ø 21.29
Median: 22
Mode: none
86. Mean: } x 5 1.9 1 1.9 1 2.5 1 2.6 1 2.8 1 3.1 1 3.5
}}}} 7
5 18.3
} 7 ø 2.61
Median: 2.6
Mode: 1.9
87. Mean: } x 5 0.9 1 1.8 1 2.5 1 3.8 1 4.2 1 5.2 1 6.7
}}}} 7
5 25.1
} 7 ø 3.59
Median: 3.8
Mode: none
88. h 5 number of hats
s 5 number of scarves
16h 1 18s 5 710 16h 1 18s 5 710
h 1 s 5 42 216h 2 16s 5 2672
2s 5 38
s 5 19
h 1 19 5 42 → h 5 23
You sold 23 hats.
Lesson 12.3
12.3 Guided Practice (pp. 810–813)
1. a2
} a1 5
27 } 81 5
1 } 3
a3
} a2 5
9 } 27 5
1 } 3
a4
} a3 5
3 } 9 5
1 } 3
a5
} a4 5
1 } 3
The series is geometric with a common ratio of 1 }
3 .
2. a2
} a1 5
2 } 1 5 2
a3
} a2 5
6 } 2 5 3
Because there is no common ratio, the sequence is not geometric.
3. a2
} a1 5
8 }
24 5 22
a3
} a2 5
216 } 8 5 22
a4
} a3 5
32 }
216 5 22
a5
} a4 5
264 } 32 5 22
The series is geometric with a common ratio of 22.
4. a1 5 3, r 5 15
} 3 5 5
an 5 a1rn 2 1
an 5 3(5)n 2 1
a8 5 3(5)8 2 1
a8 5 234,375
5. a6 5 296, r 5 2
a6 5 a1r6 2 1
296 5 a1(2)5
23 5 a1
an 5 a1(r)n 2 1
an 5 23(2)n 2 1
a8 5 23(2)8 2 1
a8 5 2384
6. a2 5 212, a4 5 23
a2 5 a1r2 2 1 → 212 5 a1r → a1 5 2 12
} r
a4 5 a1r4 2 1 → 23 5 a1r3
23 5 1 212 } r 2 r 3
23 5 212r2
1 }
4 5 r2
6 1 } 2 5 r
Chapter 12, continued
n2ws-1200-a.indd 679 6/27/06 11:32:03 AM
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680Algebra 2Worked-Out Solution Key
If r 5 1 } 2 :
a1 5 212
} 1 }
2 5 224
an 5 a1rn 2 1
an 5 224 1 1 } 2 2
n 2 1
a8 5 224 1 1 } 2 2 8 2 1
5 20.1875
If r 5 2 1 } 2 :
a1 5 212
} 2
1 } 2 5 24
an 5 a1rn 2 1
an 5 24 1 2 1 } 2 2 n 2 1
a8 5 24 1 2 1 } 2 2 7 5 20.1875
7. S8 5 a1 1 1 2 r8 }
1 2 r 2
S8 5 6 1 1 2 (22)8
} 1 2 (22)
2 S8 5 2510
8. an 5 5.02(1.059)n 2 1
a11 5 5.02(1.059)11 2 1
a11 ø 8.906
The box offi ce revenue in 2000 was about $8.91 billion.
12.3 Exercises (pp. 814–817)
Skill Practice
1. The constant ratio of consecutive terms in a geometric sequence is called the common ratio.
2. If you divide each consecutive term and get the same ratio every time, the sequence is geometric.
3. a2
} a1
5 4 } 1 5 4
a3
} a2
5 8 } 4 5 2
Because there is no common ratio, the sequence is not geometric.
4. a2
} a1
5 16
} 4 5 4
a3
} a2
5 64
} 16 5 4
a4
} a3
5 256
} 64 5 4
a5
} a4
5 1024
} 256 5 4
The sequence is geometric with a common ratio of 4.
5. a2
} a1
5 36
} 216 5 1 } 6
a3
} a2
5 6 } 36 5
1 } 6
a4
} a3
5 1 } 6
a5
} a4
5 1 } 6
The sequence is geometric with a common ratio of 1 } 6 .
6. a2
} a1
5 2 } 3 }
1 }
3 5 2
a3
} a2
5 4 } 3 }
2 }
3 5 2
a4
} a3
5 8 } 3 }
4 }
3 5 2
a5
} a4
5 16
} 3 }
8 }
3 5 2
The sequence is geometric with a common ratio of 2.
7. a2 } a1
5 1
} 1 } 2 5 2
a3 } a2
5 3 } 2 } 1 5
3 } 2
The sequence is not geometric because there is no common ratio.
8. a2 } a1
5 3 } 8 }
2 1 } 4 5 2
3 } 2
a3 } a2
5 2
3 } 16 }
3 }
8 5 2
1 } 2
The sequence is not geometric because there is no common ratio.
9. a2 } a1
5 5 } 10 5 0.5
a3 } a2
5 2.5
} 5 5 0.5
a4 } a3
5 1.25
} 2.5 5 0.5
a5 } a4
5 0.625
} 1.25 5 0.5
The sequence is geometric with a common ratio of 0.5.
10. a2
} a1 5
26 }
23 5 2
a3
} a2 5
12 }
26 5 22
The sequence is not geometric because there is no common ratio.
Chapter 12, continued
n2ws-1200-a.indd 680 6/27/06 11:32:08 AM
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681Algebra 2
Worked-Out Solution Key
11. a2
} a1 5
12 }
24 5 23
a3
} a2 5
236 } 12 5 23
a4
} a3 5
108 }
236 5 23
a5
} a4 5
2324 } 108 5 23
The sequence is geometric with a common ratio of 23.
12. a2
} a1 5
0.6 } 0.2 5 3
a3
} a2 5
1.8 } 0.6 5 3
a4
} a3 5
5.4 } 1.8 5 3
a5
} a4 5
16.2 } 5.4 5 3
The sequence is geometric with a common ratio of 3.
13. a2
} a1 5
10 }
25 5 22
a3
} a2 5
20 } 10 5 2
The sequence is not geometric because there is no common ratio.
14. a2
} a1 5
1.5 } 0.75 5 2
a3
} a2 5
2.25 } 1.5 5 1.5
The sequence is not geometric because there is no common ratio.
15. a1 5 1, r 5 2 4 } 1 5 24 16. a1 5 6, r 5
18 } 6 5 3
an 5 1(24)n 2 1 an 5 6(3)n 2 1
a7 5 (24)7 2 1 5 4096 a7 5 6(3)7 2 1 5 4374
17. a1 5 4, r 5 24
} 4 5 6 18. a1 5 7, r 5 2 35
} 7 5 25
an 5 4(6)n 2 1 an 5 7(25)n 2 1
a7 5 4(6)7 2 1 a7 5 7(25)7 2 1
5 186,624 5 109,375
19. a1 5 2, r 5 3 } 2 } 2 5
3 } 4
an 5 2 1 3 } 4 2 n 2 1
a7 5 2 1 3 } 4 2 7 2 1
5 729
} 2048
20. a1 5 3, r 5 2
6 } 5 } 3 5 2
2 } 5
an 5 3 1 2 2 } 5 2 n 2 1
a7 5 3 1 2 2 } 5 2 7 2 1
5 192
} 15,625
21. a1 5 4, r 5 2 } 4 5
1 } 2
an 5 4 1 1 } 2 2 n 2 1
a7 5 4 1 1 } 2 2 7 2 1
5 1 } 16
22. a1 5 20.3, r 5 0.6
} 20.3 5 22
an 5 20.3(22)n 2 1
a7 5 20.3(22)7 2 1 5 219.2
23. a1 5 22, r 5 20.8
} 22 5 0.4
an 5 22(0.4)n 2 1
a7 5 22(0.4)7 2 1 5 20.008192
24. a1 5 7, r 5 24.2
} 7 5 20.6
an 5 7(20.6)n 2 1
a7 5 7(20.6)7 2 1 5 0.326592
25. a1 5 5, r 5 214
} 5 5 22.8
an 5 5(22.8)n 2 1
a7 5 5(22.8)7 2 1 5 2409.45152
26. a1 5 120, r 5 180
} 120 5 1.5
an 5 120(1.5)n 2 1
a7 5 120(1.5)7 2 1 5 1366.875
27. B;
a1 5 5, r 5 20
} 5 5 4
an 5 5(4)n 2 1
28. a1 5 5, r 5 3
an 5 5(3)n 2 1
n 1 2 3 4 5 6
an 5 15 45 135 405 1215
125
n21
an
Chapter 12, continued
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682Algebra 2Worked-Out Solution Key
29. a1 5 22, r 5 6
an 5 22(6)n 2 1
n 1 2 3 4 5 6
an 22 212 272 2432 22592 215,552
1600
n1
an
30. a2 5 6, r 5 2
a2 5 a1r2 2 1
6 5 a1(2)1
3 5 a1
an 5 3(2)n 2 1
n 1 2 3 4 5 6
an 3 6 12 24 48 96
10
n21
an
31. a2 5 15, r 5 1 } 2
a2 5 a1r2 2 1
15 5 a1 1 1 } 2 2 1
30 5 a1
an 5 30 1 1 } 2 2 n 2 1
n 1 2 3 4 5 6
an 30 15 7.5 3.75 1.875 0.9375
3
n1
an
32. a5 5 1, r 5 1 } 8
a5 5 a1r5 2 1
1 5 a1 1 1 } 8 2 4
4096 5 a1
an 5 4096 1 1 } 8 2
n 2 1
n 1 2 3 4 5 6
an 4096 512 64 8 1 0.125
400
n21
an
33. a4 5 212, r 5 2 1 } 4
a4 5 a1r4 2 1
212 5 a1 1 2 1 } 4 2 3
768 5 a1
an 5 768 1 2 1 } 4 2 n 2 1
n 1 2 3 4 5 6
an 768 2192 48 212 3 20.75
100
n21
an
Chapter 12, continued
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683Algebra 2
Worked-Out Solution Key
34. a3 5 75, r 5 5
a3 5 a1r3 2 1
75 5 a1(5)2
3 5 a1
an 5 3(5)n 2 1
n 1 2 3 4 5 6
an 3 15 75 375 1875 9375
1000
n21
an
35. a2 5 8, r 5 4
a2 5 a1r2 2 1
8 5 a1(4)1
2 5 a1
an 5 2(4)n 2 1
n 1 2 3 4 5 6
an 2 8 32 128 512 2048
200
n21
an
36. a4 5 500, r 5 5
a4 5 a1r4 2 1
500 5 a1(5)3
4 5 a1
an 5 4(5)n 2 1
n 1 2 3 4 5 6
an 4 20 100 500 2500 12,500
1500
n21
an
37. In the fi rst step, the exponent should be n 2 1. an 5 a1rn 2 1
an 5 3(2)n 2 1
38. In the fi rst step, r should be raised to the (n 2 1)th power, not a1.
an 5 a1rn 2 1
an 5 3(2)n 2 1
39. a1 5 3, a3 5 12
a3 5 a1r3 2 1
12 5 3r2
4 5 r2
62 5 r
an 5 3(2)n 2 1 or an 5 3(22)n 2 1
40. a1 5 1, a5 5 625
a5 5 a1r5 2 1
625 5 1 p r4
65 5 r
an 5 1(5)n 2 1 or an 5 1(25)n 2 1
41. a1 5 2 1 } 4 , a4 5 216
a4 5 a1r4 2 1
216 5 2 1 } 4 r3
64 5 r3
4 5 r
an 5 2 1 } 4 (4)n 2 1
42. a3 5 10, a6 5 270
a3 5 a1r3 2 1 → 10 5 a1r2 → a1 5 10
} r2
a6 5 a1r6 2 1 → 270 5 a1r5
270 5 1 10 }
r2 2 r5
270 5 10r3
27 5 r3
3 5 r
a1 5 10
} 32 5 10
} 9
an 5 a1r n 2 1
an 5 10
} 9 (3)n 2 1
Chapter 12, continued
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684Algebra 2Worked-Out Solution Key
43. a2 5 240, a4 5 210
a2 5 a1r2 2 1 → 240 5 a1r → a1 5 240
} r
a4 5 a1r4 2 1 → 210 5 a1r3
210 5 1 240 } r 2 r3
210 5 240r2
1 }
4 5 r2
6 1 } 2 5 r
a1 5 240
} 1 }
2 5 280 or a1 5
240 }
2 1 } 2 5 80
an 5 a1r n 2 1 an 5 a1rn 2 1
an 5 280 1 1 } 2 2 n 2 1 an 5 80 1 2
1 }
2 2 n 2 1
44. a2 5 224, a5 5 1536
a2 5 a1r2 2 1 → 224 5 a1r → a1 5 224
} r
a5 5 a1r5 2 1 → 1536 5 a1r4
1536 5 1 224 } r 2 r4
1536 5 224r3
264 5 r3
24 5 r
a1 5 224
} 24 5 6
an 5 a1r n 2 1
an 5 6(24)n 2 1
45. a4 5 162, a7 5 4374
a4 5 a1r4 2 1 → 162 5 a1r3 → a1 5 162
} r3
a7 5 a1r7 2 1 → 4374 5 a1r6
4374 5 1 162 }
r3 2 r6
4374 5 162r3
27 5 r2
3 5 r
a1 5 162
} 33 5 6
an 5 a1r n 2 1
an 5 6(3)n 2 1
46. a3 5 7 } 4 , a5 5
7 } 16
a3 5 a1r3 2 1 → 7 }
4 5 a1r2 → a1 5
7 }
4r2
a5 5 a1r5 2 1 → 7 }
16 5 a1r4
7 }
16 5 1 7
} 4r2 2 r4
7 }
16 5
7 } 4 r2
1 }
4 5 r2
6 1 } 2 5 r
a1 5 7 } 4 }
1 6 1 } 2 2 2
5 7
an 5 a1r n 2 1
an 5 7 1 1 } 2 2
n 2 1 or an 5 7 1 2
1 } 2 2 n 2 1
47. a4 5 6, a7 5 243
} 8
a4 5 a1r4 2 1 → 6 5 a1r3 → a1 5 6 }
r3
a7 5 a1r7 2 1 → 243
} 8 5 a1r6
243
} 8 5 1 6 }
r3 2 r6
243
} 8 5 6r3
243
} 48
5 r3
3 Î}
243
} 48
5 r
3
3 Ï}
12 }
4 5 r
a1 5 6 }
1 3 3 Ï}
12 } 4 2 3
5 32
} 27
an 5 a1r n 2 1
an 5 32
} 27 1 3 3 Ï}
12 }
4 2 n 2 1
48. S10 5 a1 1 1 2 r10 }
1 2 r 2
S10 5 5 1 1 2 210 }
1 2 2 2 5 5115
49. S8 5 a1 1 1 2 r8 }
1 2 r 2
S8 5 6 1 1 2 48 }
1 2 4 2 5 131,070
Chapter 12, continued
n2ws-1200-a.indd 684 6/27/06 11:32:34 AM
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685Algebra 2
Worked-Out Solution Key
50. i 5 0
∑ 7
12 1 2 1 } 2 2 i 5
i 5 1 ∑
8
12 1 2 1 } 2 2 i 2 1
S8 5 a1 1 1 2 r8 }
1 2 r 2
S8 5 12 1 1 2 1 2 1 } 2 2 8 }
1 2 1 2 1 } 2 2 2 5
255 } 32
51. S6 5 a1 1 1 2 r6 }
1 2 r 2
S6 5 4 1 1 2 1 1 } 4 2 6 }
1 2 1 1 } 4 2 2 5
1365 } 256
52. S12 5 a1 1 1 2 r12 }
1 2 r 2
S12 5 8 1 1 2 1 3 } 2 2 12
} 1 2 1 3 } 2 2
2 5 527,345
} 256
53. i 5 0
∑ 10
(24)i 5 i 5 1
∑ 11
(24)i 2 1
S11 5 a1 1 1 2 r11 }
1 2 r 2
S11 5 1 1 1 2 (24)11
} 1 2 (24)
2 5 838,861
54. C;
S9 5 a1 1 1 2 r9 }
1 2 r 2
S9 5 2 1 1 2 39 }
1 2 3 2 5 19,682
55. Sample answer:
S5 5 100, choose r 5 2, fi nd a1.
S5 5 a1 1 1 2 r5 }
1 2 r 2
100 5 a1 1 1 2 25 }
1 2 2 2
100 5 a1(31)
100
} 31
5 a1
Series: 100
} 31
1 200
} 31 1 400
} 31 1 800
} 31 1 1600
} 31
56. a. a1 5 1, r 5 x } 1 5 x
S5 5 a1 1 1 2 r5 }
1 2 r 2
S5 5 1 1 1 2 x5 }
1 2 x 2 5
1 2 x5
} 1 2 x
b. a1 5 3x, r 5 6x3
} 3x 5 2x2
S4 5 a1 1 1 2 r4 }
1 2 r 2
S4 5 3x 1 1 2 (2x2)4
} 1 2 2x2 2
S4 5 3x 1 1 2 16x8 }
1 2 2x2 2
Problem Solving
57. a. a1 5 5, r 5 2
an 5 5(2)n 2 1
b. S4 5 a1 1 1 2 r4 }
1 2 r 2
S4 5 5 1 1 2 24 }
1 2 2 2 5 75
There are 75 skydivers in four rings.
58. a. a1 5 32, r 5 1 } 2
an 5 32 1 1 } 2 2 n 2 1
1 ≤ n ≤ 6 are the logical values because a6 5 1.
b. S6 5 a1 1 1 2 r6 }
1 2 r 2
S6 5 32 1 1 2 1 1 } 2 2 6 }
1 2 1 } 2 2 5 63
63 games are played in the tournament.
59. a. After the fi rst pass, 512 items remain, so a1 5 512
and r 5 1 } 2 .
an 5 512 1 1 } 2 2 n 2 1
b. Find n when an 5 1.
an 5 512 1 1 } 2 2 n 2 1
1 5 512 1 1 } 2 2 n 2 1
1 }
512 5 1 1 } 2 2 n 2 1
log 1 }
512 5 (n 2 1)log 1 1 }
2 2
log
1 }
512 }
log 1 }
2 1 1 5 n
10 5 n
On the tenth pass, only one term remains.
60. a. a1 5 1, r 5 8
an 5 1(8)n 2 1
an 5 8n 2 1
S8 5 a1 1 1 2 r8 }
1 2 r 2
S8 5 1 1 1 2 88 }
1 2 8 2 5 2,396,745
2,396,745 squares are removed through stage 8.
b. b1 5 1 2 1 } 9 5
8 } 9 , r 5
8 } 9
bn 5 8 } 9 1 8 }
9 2 n 2 1
b12 5 8 } 9 1 8 }
9 2 12 2 1
ø 0.2433
The remaining area of the original square after the twelfth stage is about 0.2433 square unit.
Chapter 12, continued
n2ws-1200-a.indd 685 6/27/06 11:32:39 AM
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686Algebra 2Worked-Out Solution Key
61. a. For Company A:
a1 5 20,000, d 5 1000
an 5 a1 1 (n 2 1)d
an 5 20,000 1 (n 2 1)(1000)
an 5 19,000 1 1000n
This sequence is arithmetic.
For Company B:
b1 5 20,000, r 5 1.04
bn 5 a1r n 2 1
bn 5 20,000(1.04)n 2 1
This sequence is geometric.
b.
Years of employment
Sal
ary
(do
llars
)
bn
an
19,0000
20,00021,00022,00023,00024,00025,00026,00027,000
n1 2 3 4 5 6 70
c. For Company A:
Sn 5 n 1 a1 1 an }
2 2
S20 5 20 1 20,000 1 39,000 }}
2 2 5 590,000
For Company B:
Sn 5 a1 1 1 2 rn }
1 2 r 2
S20 5 20,000 1 1 2 (1.04)20
} 1 2 1.04
2 5 595,561.57
The sum of the wages during the fi rst 20 years of employment is $590,000 for Company A and about $595,562 for Company B.
d.
For Company A
n an Sn
1 $20,000.00 $20,000.00
2 $21,000.00 $41,000.003 $22,000.00 $63,000.004 $23,000.00 $86,000.005 $24,000.00 $110,000.006 $25,000.00 $135,000.007 $26,000.00 $161,000.008 $27,000.00 $188,000.009 $28,000.00 $216,000.0010 $29,000.00 $245,000.0011 $30,000.00 $275,000.0012 $31,000.00 $306,000.0013 $32,000.00 $338,000.0014 $33,000.00 $371,000.0015 $34,000.00 $405,000.0016 $35,000.00 $440,000.0017 $36,000.00 $476,000.0018 $37,000.00 $513,000.0019 $38,000.00 $551,000.0020 $39,000.00 $590,000.00
For Company B
n Sn
1 $20,000.00
2 $40,800.003 $62,432.004 $84,929.285 $108,326.456 $132,659.517 $157,965.898 $184,284.539 $211,655.9110 $240,122.1411 $269,727.0312 $300,516.1113 $332,536.7514 $365,838.2215 $400,471.7516 $436,490.6217 $473,950.2518 $512,908.2619 $553,424.5920 $595,561.57
The total amount earned by Company B is greater than the amount earned by Company A after 19 years.
62. a1 5 2000, r 5 1.05
S30 5 a1 1 1 2 r30 }
1 2 r 2
S30 5 2000 1 1 2 1.0530 }
1 2 1.05 2 ø 132,877.70
You will have $132,877.70 in the IRA after your last deposit.
Mixed Review
63.
10 221
13
29
12
56
64.
2 3 4 51023 22 21
472 6
65.
2 31023 22 212627 25 24
1142 8 2.721.8
66. 4 }
1 1 x 5 9 67.
3 }
1 2 x 5 10
4 5 9(1 1 x) 3 5 10(1 2 x)
4 5 9 1 9x 3 5 10 2 10x
25 5 9x 27 5 210x
2 5 } 9 5 x x 5
7 } 10
Chapter 12, continued
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687Algebra 2
Worked-Out Solution Key
68. 2 }
3x 5
5 } x 1 4
2(x 1 4) 5 5(3x)
2x 1 8 5 15x
8 5 13x
8 }
13 5 x
69. x }
x 2 6 5
x } 3
x(3) 5 x(x 2 6)
3x 5 x2 2 6x
0 5 x2 2 9x
0 5 x(x 2 9)
x 5 0 or x 5 9
70. 2 18
} x 2 x 5 11
x 1 2 18
} x 2 x 2 5 x(11)
218 2 x2 5 11x
0 5 x2 1 11x 1 18
0 5 (x 1 2)(x 1 9)
x 5 22 or x 5 29
71. x 1 16 5 x2
} x 2 8
(x 2 8)(x 1 16) 5 x2
x2 1 16x 2 8x 2 128 5 x2
8x 2 128 5 0
8x 5 128
x 5 16
72. i 5 2
∑ 8
(i2 1 1) 5 (22 1 1) 1 (32 1 1) 1 (42 1 1) 1 (52 1 1) 1 (62 1 1) 1 (72 1 1) 1 (82 1 1) 5 5 1 10 1 17 1 26 1 37 1 50 1 65
5 210
73. i 5 3
∑ 11
6i 5 6(3) 1 6(4) 1 6(5) 1 6(6) 1 6(7)
1 6(8) 1 6(9) 1 6(10) 1 6(11)
5 18 1 24 1 30 1 36 1 42 1 48 1 54 1 60 1 66
5 378
74. i 5 1
∑ 6
(5 2 i) 5 (5 2 1) 1 (5 2 2) 1 (5 2 3)
1 (5 2 4) 1 (5 2 5) 1 (5 2 6)
5 4 1 3 1 2 1 1 1 0 1 (21)
5 9
75. i 5 7
∑ 15
(3i 1 4) 5 (3(7) 1 4) 1 (3(8) 1 4)
1 (3(9) 1 4) 1 . . . 1 (3(14) 1 4)
1 (3(15) 1 4)
5 25 1 28 1 31 1 34 1 37 1 40
1 43 1 46 1 49
5 333
76. i 5 5
∑ 12
(22i 1 1) 5 (22(5) 1 1) 1 (22(6) 1 1) 1 . . .
1 (22(11) 1 1) 1 (22(12) 1 1)
5 29 2 11 2 13 2 15 2 17
2 19 2 21 2 23
5 2128
77. i 5 4
∑ 9
4i2 5 4(4)2 1 4(5)2 1 4(6)2 1 4(7)2 1 4(8)2 1 4(9)2
5 64 1 100 1 144 1 196 1 256 1 324
5 1084
Quiz 12.1–12.3 (p. 817)
1. Given terms: 1, 3, 5, 7, . . .
Rewritten terms: 2(1) 2 1, 2(2) 2 1, 2(3) 2 1, 2(4) 2 1
Next term: 2(5) 2 1 5 9
Rule for nth term: an 5 2n 2 1
2. Given terms: 25, 10, 215, 20, . . .
Rewritten terms: (21)1(5(1)), (21)2(5(2)), (21)3(5(3)), (21)4(5(4))
Next term: (21)5(5(5)) 5 225
Rule for nth term: an 5 (21)n(5n)
3. Given terms: 1 }
20 , 2 }
30 , 3 }
40 , 4 }
50 , . . .
Rewritten terms: 1 }
(1 1 1)(10) ,
2 }
(2 1 1)(10) ,
3 }
(3 1 1)(10) ,
4 }
(4 1 1)(10)
Next term: 5 }
(5 1 1)(10) 5
5 } 60
Rule for nth term: an 5 n }
10(n 1 1)
4. Given terms: 4, 16, 64, 256, . . .
Rewritten terms: 41, 42, 43, 44
Next term: 45 5 1024
Rule for nth term: an 5 4n
5. Given terms: 2, 6, 12, 20, . . .
Rewritten terms: 1(1 1 1), 2(2 1 1), 3(3 1 1), 4(4 1 1)
Next term: 5(5 1 1) 5 30
Rule for nth term: an 5 n(n 1 1)
6. Given terms: 9, 36, 81, 144, . . .
Rewritten terms: 9(12), 9(22), 9(32), 9(42) Next term: 9(52) 5 225
Rule for nth term: an 5 9n2 5 (3n)2
7. i 5 1
∑ 4
2i3 5 2(13) 1 2(23) 1 2(33) 1 2(43) 5 2 1 16 1 54 1 128
5 200
8. k 5 1
∑ 5
(k2 1 3) 5 (12 1 3) 1 (22 1 3) 1 (32 1 3) 1 (42 1 3) 1 (52 1 3) 5 4 1 7 1 12 1 19 1 28
5 70
Chapter 12, continued
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688Algebra 2Worked-Out Solution Key
9. n 5 2
∑ 6
1 }
n 2 1 5
1 } 2 2 1 1
1 } 3 2 1 1
1 } 4 2 1 1
1 } 5 2 1 1
1 } 6 2 1
5 1 1 1 } 2 1
1 } 3 1
1 } 4 1
1 } 5
5 137
} 60
10. a1 5 1, d 5 6
an 5 a1 1 (n 2 1)d
an 5 1 1 (n 2 1)(6)
an 5 25 1 6n
a15 5 25 1 6(15) 5 85
Sn 5 n 1 a1 1 a15 } 2 2
S15 5 15 1 1 1 85 }
2 2 5 645
11. a1 5 1 } 2 , d 5
3 } 2
an 5 a1 1 (n 2 1)d
an 5 1 } 2 1 (n 2 1) 1 3 }
2 2
an 5 21 1 3 } 2 n
a15 5 21 1 3 } 2 (15) 5
43 } 2
Sn 5 n 1 a1 1 an } 2 2
S15 5 15 1 1 }
2 1
43 } 2 }
2 2 5 165
12. a1 5 5, d 5 23
an 5 a1 1 (n 2 1)d
an 5 5 1 (n 2 1)(23)
an 5 8 2 3n
a15 5 8 2 3(15) 5 237
Sn 5 n 1 a1 1 an } 2 2
S15 5 15 1 5 1 (237) }
2 2 5 2240
13. a1 5 2, r 5 8 } 2 5 4
an 5 a1rn 2 1
an 5 2(4)n 2 1
a15 5 2(4)15 2 1 5 536,870,912
Sn 5 a1 1 1 2 rn }
1 2 r 2
S15 5 2 1 1 2 415 }
1 2 4 2 5 715,827,882
14. a1 5 2, r 5 4 } 3 } 2 5
2 } 3
an 5 a1rn 2 1
an 5 2 1 2 } 3 2
n 2 1
a15 5 2 1 2 } 3 2
15 2 1 5
32,768 } 4,782,969
Sn 5 a1 1 1 2 rn }
1 2 r 2
S15 5 2 1 1 2 1 2 } 3 2 15
} 1 2 1 2 } 3 2
2 ø 5.986
15. a1 5 23, r 5 15
} 23 5 25
an 5 a1rn 2 1
an 5 23(25)n 2 1
a15 5 23(25)15 2 1 5 218,310,546,875
Sn 5 a1 1 1 2 rn }
1 2 r 2
S15 5 23 1 1 2 (25)15
} 1 2 (25)
2 5 215,258,789,063
16. a1 5 2057, r 5 1.06
an 5 a1rn 2 1
an 5 2057(1.06)n 2 1
In 2002, n 5 8.
a8 5 2057(1.06)8 2 1 ø 3092.97
In 2002, the average tuition at a public college was about $3092.97.
Mixed Review of Problem Solving (p. 818)
1. a. a1 5 45,000, r 5 1.035
an 5 a1rn 2 1
an 5 45,000(1.035)n 2 1
b. a5 5 45,000(1.035)5 2 1 ø 51,638.54
During your 5th year of employment, your salary will be $51,638.54.
c. Sn 5 a1 1 1 2 rn }
1 2 r 2
S30 5 45,000 1 1 2 (1.035)30
} 1 2 1.035
2 ø 2,323,020.48
After 30 years, you will have earned a total of $2,323,020.48.
2. a. A 5 πr2
a1 5 π(12) 5 π
a2 5 π(22) 2 π(12) 5 3π
a3 5 π(32) 2 π(22) 5 5π
a1 5 π, d 5 2π
an 5 π 1 (n 2 1)(2π) 5 (2n 2 1)π
b. i 5 1
∑ n
(2i 2 1)π
Chapter 12, continued
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689Algebra 2
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c. For n 5 1: i 5 1
∑ 1
(2i 2 1)π 5 π
For n 5 2: i 5 1
∑ 2
(2i 2 1)π 5 π 1 3π 5 4π
For n 5 4: i 5 1
∑ 4
(2i 2 1)π 5 π 1 3π 1 5π 1 7π
5 16π
For n 5 8:
i 5 1
∑ 8
(2i 2 1)π 5 π 1 3π 1 5π 1 7π 1 9π
1 11π 1 13π 1 15π 5 64π
When the number of rings is doubled, the total area is quadrupled.
3. Tables placed together by short edges:
Sequence: 6, 10, 14, 18 . . .
a1 5 6, d 5 4
an 5 6 1 (n 2 1)(4) 5 2 1 4n
Tables placed together by long edges:
Sequence: 6, 8, 10, 12 . . .
a1 5 6, d 5 2
an 5 6 1 (n 2 1)(2) 5 4 1 2n
When the tables are connected by the short edges, 2 1 4n 2 (4 1 2n), or 22 1 2n, more people can be seated.
4. Sample answer:
n 5 8, Sn 5 70
Sn 5 n 1 a1 1 an } 2 2
70 5 8 1 a1 1 a8 } 2 2
17.5 5 a1 1 a8
Choose a1 5 1.75 and a8 5 15.75.
an 5 a1 1 (n 2 1)d
a8 5 a1 1 (8 2 1)d
15.75 5 1.75 1 7d
14 5 7d
2 5 d
A series is i 5 1
∑ 8
(2i 2 0.25). So, an 5 1.75 1 (n 2 1)(2),
or an 5 20.25 1 2n.
5. a1 5 15, d 5 21
an 5 a1 1 (n 2 1)d
an 5 15 1 (n 2 1)(21)
an 5 16 2 n
6 5 16 2 n
n 5 10
Sn 5 n 1 a1 1 an }
2 2
S10 5 10 1 15 1 6 }
2 2 5 105
There are 105 pieces of chalk in the pile.
6. a1 5 9, d 5 7
an 5 a1 1 (n 2 1)d
an 5 9 1 (n 2 1)(7)
an 5 2 1 7n
The height at the top of the 10th stair is 2 1 7(10) 5 72 inches. To fi nd the height of the bottom of the nth stair, subtract the height of the nth stair from an. So, an 5 2 1 7n 2 7 5 25 1 7n.
7. a. The sequence is geometric because each term is half of the previous term.
b. a1 5 66, r 5 1 } 2
an 5 a1rn 2 1
an 5 66 1 1 } 2 2 n 2 1
c.
n 1 2 3 4 5 6 7
an 66 33 16.5 8.25 4.125 2.0625 1.03125
7
n
an
1
The points lie on an exponential decay curve.
d. an < 1
66 1 1 } 2 2 n 2 1
< 1
1 1 } 2 2 n 2 1
< 1 }
66
(n 2 1) log 1 1 } 2 2 < log
1 }
66
n 2 1 < log
1 }
66 }
log 1 }
2
n < 7.044
Because n represents the number of two-hour intervals, there will be less than one gram of Platinum-197 after about 14 hours.
Chapter 12, continued
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690Algebra 2Worked-Out Solution Key
8. Sample answer:
Choose a geometric series with fi ve terms: 1 1 2 1 4 1 8 1 16 5 31.
Next, fi nd an arithmetic series with fi ve terms and a sum of 31.
Sn 5 1 a1 1 an } 2 2
31 5 5 1 a1 1 a5 }
2 2
12.4 5 a1 1 a5
Choose a1 5 2.2 and a5 5 10.2.
an 5 a1 1 (n 2 1)d
10.2 5 2.2 1 4d
8 5 4d
2 5 d
So, an 5 2.2 1 (n 2 1)(2) 5 0.2 1 2n and the series is 2.2 1 4.2 1 6.2 1 8.2 1 10.2 5 31.
Lesson 12.4
Investigating Algebra Activity 12.4 (p. 819)
Step 4. The next three areas are 1 }
8 ,
1 }
16 , and
1 }
32 . These areas
form a geometric sequence because there is a common
ratio of 1 }
2 .
Step 5.
Number of pieces
1 2 3
Combined area
1 } 2
1 }
2 1
1 } 4 5
3 } 4
3 }
4 1
1 } 8 5
7 } 8
Number of pieces
4 5
Combined area
7 } 8 1
1 } 16 5
15 } 16
15 }
16 1
1 } 32 5
31 } 32
1. The total area appears to be approaching one square unit.
2. An 5 A1 1 1 2 rn }
1 2 r 2
An 5 1 } 2 1 1 2 1 1 } 2 2 n
} 1 2
1 } 2 2 5 1 2 1 1 } 2 2 n
As n gets large, An gets close to 1. For example, when n 5 10, An 5 0.9990234375, and when n 5 100, An ø 1.
12.4 Guided Practice (pp. 821–822)
1. S1 5 2 } 5 5 0.40
S2 5 2 } 5 1
4 } 25 5 0.56
S3 5 2 } 5 1
4 } 25 1
8 } 125 ø 0.62
S4 5 2 } 5 1
4 } 25 1
8 } 125 1
16 } 625 ø 0.65
S5 5 2 } 5 1
4 } 25 1
8 } 125 1
16 } 625 1
32 } 3125 ø 0.66
As n increases, Sn appears to approach 2 }
3 .
n1
0.1
Sn
2. a1 5 1, r 5 2 1 } 2
S 5 a1 } 1 2 r 5
1 }
1 2 1 2 1 } 2 2
5 2 } 3
3. r 5 5 } 4
Because 5 }
4 ≥ 1, the series has no sum.
4. a1 5 3, r 5 3 } 4 } 3 5
1 } 4
S 5 a1 } 1 2 r 5
3 }
1 2 1 } 4 5 4
5. a1 5 10, r 5 0.8
d 5 a1 } 1 2 r 5
10 } 1 2 0.8 5 50
The pendulum swings a total distance of 50 inches.
6. 0.555. . . 5 5(0.1) 1 5(0.1)2 1 5(0.1)3 1 . . .
5 a1 } 1 2 r
5 5(0.1)
} 1 2 0.1
5 0.5
} 0.9
5 5 } 9
7. 0.727272. . . 5 72(0.01) 1 72(0.01)2 1 72(0.01)3 1 . . .
5 a1 } 1 2 r
5 72(0.01)
} 1 2 0.01
5 0.72
} 0.99
5 8 } 11
Chapter 12, continued
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691Algebra 2
Worked-Out Solution Key
8. 0.131313. . . 5 13(0.01) 1 13(0.01)2 1 13(0.01)3 1 . . .
5 a1 } 1 2 r
5 13(0.01)
} 1 2 0.01
5 0.13
} 0.99
5 13
} 99
12.4 Exercises (pp. 823–825)
Skill Practice
1. The sum Sn of the fi rst n terms of an infi nite series is called a partial sum.
2. If r < 1, the series has a sum.
3. S1 5 1 } 2 5 0.5
S2 5 1 } 2 1
1 } 6 ø 0.67
S3 5 1 } 2 1
1 } 6 1
1 } 18 ø 0.72
S4 5 1 } 2 1
1 } 6 1
1 } 18 1
1 } 54 ø 0.74
S5 5 1 } 2 1
1 } 6 1
1 } 18 1
1 } 54 1
1 } 162 ø 0.75
Sn appears to be approaching 3 }
4 .
0.1
x
y
1
4. S1 5 2 } 3 ø 0.67
S2 5 2 } 3 1
1 } 3 5 1
S3 5 2 } 3 1
1 } 3 1
1 } 6 ø 1.17
S4 5 2 } 3 1
1 } 3 1
1 } 6 1
1 } 12 5 1.25
S5 5 2 } 3 1
1 } 3 1
1 } 6 1
1 } 12 1
1 } 24 ø 1.29
Sn appears to be approaching 1 1 }
3 .
0.15
x
y
1
5. S1 5 4
S2 5 4 1 12
} 5 5 6.4
S3 5 4 1 12
} 5 1 36
} 25 5 7.84
S4 5 4 1 12
} 5 1 36
} 25 1 108
} 125 ø 8.70
S5 5 4 1 12
} 5 1 36
} 25 1 108
} 125 1 324
} 625 ø 9.22
Sn appears to be approaching 10.
1
x
y
1
6. S1 5 1 } 4 5 0.25
S2 5 1 } 4 1
5 } 4 5 1.5
S3 5 1 } 4 1
5 } 4 1
25 } 4 5 7.75
S4 5 1 } 4 1
5 } 4 1
25 } 4 1
125 } 4 5 39
S5 5 1 } 4 1
5 } 4 1
25 } 4 1
125 } 4 1
625 } 4 5 195.25
As n increases, Sn also increases. The series has no sum.
20
x
y
1
7. a1 5 8, r 5 1 } 5
S 5 a1 } 1 2 r 5
8 }
1 2 1 } 5 5 10
8. r 5 3 } 2
Because 3 } 2 ≥ 1, the series has no sum.
9. r 5 5 } 3
Because 5 } 3 ≥ 1, the series has no sum.
10. a1 5 11
} 3 , r 5 3 } 8
S 5 a1 } 1 2 r 5
11
} 3 }
1 2 3 } 8 5
88 } 15
Chapter 12, continued
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692Algebra 2Worked-Out Solution Key
11. a1 5 2, r 5 1 } 6
S 5 a1 } 1 2 r 5
2 }
1 2 1 } 6 5
12 } 5
12. a1 5 25, r 5 2 } 5
S 5 a1 } 1 2 r 5
25 }
1 2 2 } 5 5 2
25 } 3
13. a1 5 7, r 5 2 8 } 9
S 5 a1 } 1 2 r 5
7 }
1 2 1 2 8 } 9 2 5
63 } 17
14. r 5 2 10
} 3
Because 2 10
} 3 ≥ 1, the series has no sum.
15. r 5 4
Because 4 ≥ 1, the series has no sum.
16. a1 5 22, r 5 2 1 } 4
S 5 a1 } 1 2 r 5
22 }
1 2 1 2 1 } 4 2 5 2
8 } 5
17. a1 5 1, r 5 2 3 } 7
S 5 a1 } 1 2 r 5
1 }
1 2 1 2 3 } 7 2 5
7 } 10
18. r 5 3
Because 3 ≥ 1, the series has no sum.
19. For this series, r 5 7 } 2 .
Because 7 }
2 ≥ 1, the series has no sum.
20. a1 5 2 1 } 8 , r 5
2 1 } 12 }
2 1 } 8 5
2 } 3
S 5 a1 } 1 2 r 5
2 1 } 8 }
1 2 2 } 3 5 2
3 } 8
21. a1 5 2 } 3 , r 5
2 2 } 9 }
2 } 3 5 2
1 } 3
S 5 a1 } 1 2 r 5
2 } 3 }
1 2 1 2 1 } 3 2 5
1 } 2
22. a1 5 4 } 15 , r 5
4 } 9 }
4 }
15 5
5 } 3
Because 5 }
3 ≥ 1, the series has no sum.
23. a1 5 3, r 5 5 } 2 } 3 5
5 } 6
S 5 a1 } 1 2 r 5
3 }
1 2 5 } 6 5 18
24. 0.222. . . 5 2(0.1) 1 2(0.1)2 1 2(0.1)3 1 . . .
5 a1 } 1 2 r
5 2(0.1)
} 1 2 0.1
5 0.2
} 0.9
5 2 } 9
25. 0.444. . . 5 4(0.1) 1 4(0.1)2 1 4(0.1)3 1 . . .
5 a1 } 1 2 r
5 4(0.1)
} 1 2 0.1
5 0.4
} 0.9
5 4 } 9
26. 0.161616. . . 5 16(0.01) 1 16(0.01)2 1 16(0.01)3 1 . . .
5 a1 } 1 2 r
5 16(0.01)
} 1 2 0.01
5 0.16
} 0.99
5 16
} 99
27. 0.625625625. . . 5 625(0.001) 1 625(0.001)2
1 625(0.001)3 1 . . .
5 a1 } 1 2 r
5 625(0.001)
} 1 2 0.001
5 0.625
} 0.999
5 625
} 999
28. 32.3232. . . 5 32 1 32(0.01) 1 32(0.01)2 1 . . .
5 a1 } 1 2 r
5 32 } 1 2 0.01
5 32
} 0.99
5 3200
} 99
29. 130.130130. . . 5 130 1 130(0.001)
1 130(0.001)2 1 . . .
5 a1 } 1 2 r
5 130 } 1 2 0.001
5 130
} 0.999
5 130,000
} 999
Chapter 12, continued
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693Algebra 2
Worked-Out Solution Key
Chapter 12, continued
30. 0.090909. . . 5 9(0.01) 1 9(0.01)2 1 9(0.01)3 1 . . .
5 a1 } 1 2 r
5 9(0.01)
} 1 2 0.01
5 0.09
} 0.99
5 1 } 11
31. 0.2777. . . 5 0.2 1 7(0.1)2 1 7(0.1)3 1 7(0.1)4 1 . . .
5 0.2 1 a1 } 1 2 r
5 0.2 1 7(0.1)2
} 1 2 0.1
5 0.2 1 0.07
} 0.9
5 18
} 90 1 7 } 90 5
5 } 18
32. C;
18.1818. . . 5 18 1 18(0.01) 1 18(0.01)2 1 . . .
5 18 } 1 2 0.01
5 1800
} 0.99
5 1800
} 99
5 200
} 11
33. 0.999. . . 5 9(0.1) 1 9(0.1)2 1 9(0.1)3 1 . . .
5 a1 } 1 2 r
5 9(0.1)
} 1 2 0.1
5 0.9
} 0.9
5 1
34. Sample answer:
a. Choose r 5 1 } 2 .
S 5 a1 } 1 2 r
5 5 a1 }
1 2 1 } 2
5 }
2 5 a1
The fi rst series is i 5 1
∑ `
5 }
2 1 1 }
2 2 i 2 1
.
b. Choose r 5 1 } 5 .
S 5 a1 } 1 2 r
5 5 a1 }
1 2 1 } 5
4 5 a1
The fi rst series is i 5 1
∑ `
4 1 1 } 5 2 i 2 1
.
35. a1 5 1, r 5 4x
} 1 5 4x
If the series has a sum, 2 1 } 4 < x <
1 }
4 .
S 5 a1 } 1 2 r 5
1 } 1 2 4x
36. a1 5 6, r 5 3 } 2 x } 6 5
1 } 4 x
If the series has a sum, 24 < x < 4.
S 5 a1 } 1 2 r 5
6 }
1 2 1 } 4 x
Problem Solving
37. d 5 14 1 14(0.8) 1 14(0.8)2 1 . . .
5 a1 } 1 2 r
5 14 } 1 2 0.8
5 70
The person swings a total distance of 70 feet.
38. S 5 350,000 1 350,000(0.88) 1 350,000(0.88)2 1 . . .
5 a1 } 1 2 r
5 350,000
} 1 2 0.88
ø 2,916,666.67
The maximum amount of profi t the company can make is $2,916,666.67.
39. D;
S 5 345 1 345(0.783) 1 345(0.783)2 1 . . .
5 a1 } 1 2 r
5 345 } 1 2 0.783
ø 1589.86
The company will ship a total of about 1.59 billion cassettes.
40. Distance: a1 5 20, r 5 10
} 20 5 1 } 2
Sd 5 a1 } 1 2 r
Sd 5 20 }
1 2 1 } 2 5 40
Time: a1 5 1, r 5 0.5
} 1 5 0.5
St 5 a1 } 1 2 r
St 5 1 } 1 2 0.5 5 2
Because both series have fi nite sums, Archilles catches up to the tortoise in 2 seconds after 40 feet is traveled.
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694Algebra 2Worked-Out Solution Key
41. a. The ball bounces 6 1 6, or 12 feet between the fi rst and second bounce, and 4.5 1 4.5, or 9 feet between the second and third bounce.
b. a1 5 12, r 5 0.75
an 5 a1rn 2 1
an 5 12(0.75)n 2 1
Series: i 5 1
∑ `
12(0.75)i 2 1
c. S 5 a1 } 1 2 r 5
12 } 1 2 0.75 5 48
The ball travels a total distance of 48 1 8 5 56 feet.
d. a1 5 2(0.75)h, r 5 0.75
Total distance 5 distance from each bounce
1 original distance
5 a1 } 1 2 r 1 h
5 2(0.75)h
} 1 2 0.75 1 h
5 6h 1 h
5 7h
If the ball is dropped from a distance of h feet, it travels a total distance of 7h feet.
42. a. a1 5 1 } 4 , a2 5
3 } 16 , r 5
3 } 16 }
1 }
4 5
3 } 4
an 5 a1rn 2 1
an 5 1 } 4 1 3 } 4 2
n 2 1
b. n 5 1
∑ `
an 5 n 5 1
∑ `
1 }
4 1 3 }
4 2
n 2 1
5 1 } 4 }
1 2 3 } 4
5 1
This answer means that eventually, one square unit of area will be removed from the triangle, so no area will remain.
Mixed Review
43. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.82 5 0.32 1 P(B) 2 0.11
0.61 5 P(B)
P(B) 5 61%
44. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.6 5 P(A) 1 0.17 2 0.03
0.46 5 P(A)
45. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.5 5 0.2 1 0.4 2 P(A and B)
P(A and B) 5 0.1
46. d 5 5, a1 5 2
an 5 a1 1 (n 2 1)d
an 5 2 1 (n 2 1)(5) 5 23 1 5n
47. d 5 8, a1 5 227
an 5 a1 1 (n 2 1)d
an 5 227 1 (n 2 1)(8) 5 235 1 8n
48. d 5 218, a8 5 72
an 5 a1 1 (n 2 1)d
a8 5 a1 1 (8 2 1)d
72 5 a1 1 7(218)
198 5 a1
an 5 198 1 (n 2 1)(218) 5 216 2 18n
49. d 5 27, a7 5 28
an 5 a1 1 (n 2 1)d
a7 5 a1 1 (7 2 1)d
28 5 a1 1 6(27)
34 5 a1
an 5 34 1 (n 2 1)(27) 5 41 2 7n
50. d 5 6.5, a5 5 92
an 5 a1 1 (n 2 1)d
a5 5 a1 1 (5 2 1)d
92 5 a1 1 4(6.5)
66 5 a1
an 5 66 1 (n 2 1)(6.5) 5 59.5 1 6.5n
51. d 5 21.5, a9 5 4
an 5 a1 1 (n 2 1)d
a9 5 a1 1 (9 2 1)d
4 5 a1 1 8(21.5)
16 5 a1
an 5 16 1 (n 2 1)(21.5) 5 17.5 2 1.5n
52. r 5 2.5, a3 5 25 53. r 5 23, a2 5 218
an 5 a1rn 2 1 an 5 a1rn 2 1
a3 5 a1r3 2 1 a2 5 a1r2 2 1
25 5 a1(2.5)2 218 5 a1(23)1
4 5 a1 6 5 a1
an 5 4(2.5)n 2 1 an 5 6(23)n 2 1
54. r 5 20.25, a5 5 40.5
an 5 a1rn 2 1
a5 5 a1r5 2 1
240.5 5 a1(0.25)4
210,368 5 a1
an 5 210,368(0.25)n 2 1
Chapter 12, continued
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695Algebra 2
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55. r 5 6, a4 5 24 56. r 5 0.75, a6 5 30
an 5 a1rn 2 1 an 5 a1r n 2 1
a4 5 a1r4 2 1 a6 5 a1r6 2 1
24 5 a1(6)3 30 5 a1(0.75)5
1 }
9 5 a1
10,240 }
81 5 a1
an 5 1 } 9 (6)n 2 1 an 5
10,240 } 81 (0.75)n 2 1
57. r 5 4, a4 5 128
an 5 a1rn 2 1
a4 5 a1r4 2 1
128 5 a1(4)3
2 5 a1
an 5 2(4)n 2 1
Lesson 12.5
Investigating Algebra Activity 12.5 (p. 826)
1. Sequence: 4; 28; 196; 1372; 9604; 67,228; 470,596; 3,294,172
The sequence is geometric with a common ratio of 7.
2. a1 5 15, d 5 11 2 15 5 24
an 5 an 2 1 2 4
3. a1 5 81, r 5 27
} 81 5 1 } 3
an 5 1 } 3 an 2 1
4. Arithmetic sequences: an 5 an 2 1 1 d
Geometric sequences: an 5 ran 2 1
12.5 Guided Practice (pp. 828–830)
1. a1 5 3
a2 5 a1 2 7 5 3 2 7 5 24
a3 5 a2 2 7 5 24 2 7 5 211
a4 5 a3 2 7 5 211 2 7 5 218
a5 5 a4 2 7 5 218 2 7 5 225
2. a0 5 162
a1 5 0.5a0 5 0.5(162) 5 81
a2 5 0.5a1 5 0.5(81) 5 40.5
a3 5 0.5a2 5 0.5(40.5) 5 20.25
a4 5 0.5a3 5 0.5(20.25) 5 10.125
3. a0 5 1
a1 5 a0 1 1 5 1 1 1 5 2
a2 5 a1 1 2 5 2 1 2 5 4
a3 5 a2 1 3 5 4 1 3 5 7
a4 5 a3 1 4 5 7 1 4 5 11
4. a1 5 4
a2 5 2a1 2 1 5 2(4) 2 1 5 7
a3 5 2a2 2 1 5 2(7) 2 1 5 13
a4 5 2a3 2 1 5 2(13) 2 1 5 25
a5 5 2a4 2 1 5 2(25) 2 1 5 49
5. a1 5 21, r 5 14
} 2 5 7
an 5 r p an 2 1
an 5 7an 2 1
6. a1 5 19, d 5 13 2 19 5 26
an 5 an 2 1 1 d
an 5 an 2 1 2 6
7. a1 5 11, d 5 22 2 11 5 11
an 5 an 2 1 1 d
an 5 an 2 1 1 11
8. a1 5 324, r 5 108
} 324 5 1 } 3
an 5 r p an 2 1
an 5 1 } 3 an 2 1
9. a1 5 1
a2 5 2
a3 5 2 5 a1 p a2
a4 5 4 5 a2 p a3
a1 5 1, a2 5 2, an 5 (an 2 2)(an 2 1)
10. a1 5 50,000, an 5 0.7an 2 1 1 5000
The number of members stabilizes at about 16,667 members.
11. f (x) 5 4x 2 3, x0 5 2
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (2) 5 f (5) 5 f (17)
5 4(2) 2 3 5 4(5) 2 3 5 4(17) 2 3
5 5 5 17 5 65
12. f (x) 5 x2 2 5, x0 5 21
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (21) 5 f (24) 5 f (11)
5 (21)2 2 5 5 (24)2 2 5 5 (11)2 2 5
5 24 5 11 5 116
12.5 Exercises (pp. 830–833)
Skill Practice
1. The repeated composition of a function with itself is called iteration.
2. An explicit rule gives the value based on the position of the term in the sequence. A recursive rule gives the value based on the previous term in the sequence.
Chapter 12, continued
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696Algebra 2Worked-Out Solution Key
3. a1 5 1
a2 5 a1 1 3 5 1 1 3 5 4
a3 5 a2 1 3 5 4 1 3 5 7
a4 5 a3 1 3 5 7 1 3 5 10
a5 5 a4 1 3 5 10 1 3 5 13
4. a0 5 4
a1 5 2a0 5 2(4) 5 8
a2 5 2a1 5 2(8) 5 16
a3 5 2a2 5 2(16) 5 32
a4 5 2a3 5 2(32) 5 64
5. a1 5 21
a2 5 a1 2 5 5 21 2 5 5 26
a3 5 a2 2 5 5 26 2 5 5 211
a4 5 a3 2 5 5 211 2 5 5 216
a5 5 a4 2 5 5 216 2 5 5 221
6. a0 5 3
a1 5 a0 2 12 5 3 2 1 5 2
a2 5 a1 2 22 5 2 2 4 5 22
a3 5 a2 2 32 5 22 2 9 5 211
a4 5 a3 2 42 5 211 2 16 5 227
7. a1 5 2
a2 5 a12 1 1 5 22 1 1 5 5
a3 5 a22 1 1 5 52 1 1 5 26
a4 5 a32 1 1 5 262 1 1 5 677
a5 5 a42 1 1 5 6772 1 1 5 458,330
8. a0 5 4
a1 5 (a0)2 2 10 5 42 2 10 5 6
a2 5 (a1)2 2 10 5 62 2 10 5 26
a3 5 (a2)2 2 10 5 262 2 10 5 666
a4 5 (a3)2 2 10 5 6662 2 10 5 443,546
9. a1 5 2
a2 5 22 1 3(2) 2 a1 5 4 1 6 2 2 5 8
a3 5 32 1 3(3) 2 a2 5 9 1 9 2 8 5 10
a4 5 42 1 3(4) 2 a3 5 16 1 12 2 10 5 18
a5 5 52 1 3(5) 2 a4 5 25 1 15 2 18 5 22
10. a0 5 2
a1 5 4
a2 5 a1 2 a0 5 4 2 2 5 2
a3 5 a2 2 a1 5 2 2 4 5 22
a4 5 a3 2 a2 5 22 2 2 5 24
11. a1 5 2
a2 5 3
a3 5 a2 p a1 5 3 p 2 5 6
a4 5 a3 p a2 5 6 p 3 5 18
a5 5 a4 p a3 5 18 p 6 5 108
12. A;
a1 5 1
a2 5 4
a3 5 a2 p a1 5 4 p 1 5 4
a4 5 a3 p a2 5 4 p 4 5 16
13. a1 5 21, d 5 14 2 21 5 27
an 5 an 2 1 1 d 5 an 2 1 2 7
14. a1 5 3, r 5 12
} 3 5 4
an 5 ran 2 1 5 4an 2 1
15. a1 5 4, r 5 2 12
} 4 5 23
an 5 ran 2 1 5 23an 2 1
16. a1 5 1, d 5 8 2 1 5 7
an 5 an 2 1 1 d 5 an 2 1 1 7
17. a1 5 44, r 5 11
} 44 5 1 } 4
an 5 ran 2 1 5 1 } 4 an 2 1
18. a1 5 1, a2 5 4
a3 5 5 5 a2 1 a1
a4 5 9 5 a3 1 a2
a1 5 1, a2 5 4, an 5 an 2 1 1 an 2 2
19. a1 5 54, d 5 43 2 54 5 211
an 5 an 2 1 1 d 5 an 2 1 2 11
20. a1 5 3, a2 5 5
a3 5 15 5 a2 p a1
a4 5 75 5 a3 p a2
a1 5 3, a2 5 5, an 5 an 2 1 p an 2 2
21. a1 5 16, a2 5 9
a3 5 7 5 a1 2 a2
a4 5 2 5 a2 2 a3
a1 5 16, a2 5 9, an 5 an 2 2 2 an 2 1
22. The previous terms must be defi ned fi rst.
a1 5 5, a2 5 2, an 5 an 2 2 2 an 2 1
23. The rule does not work for all the terms in the sequence.
a1 5 5, a2 5 2, an 5 an 2 2 2 an 2 1
24. f (x) 5 3x 2 2, x0 5 2
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (2) 5 f (4) 5 f (10)
5 3(2) 2 2 5 3(4) 2 2 5 3(10) 2 2
5 4 5 10 5 28
25. f (x) 5 5x 1 6, x0 5 22
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (22) 5 f (24) 5 f (214)
5 5(22) 1 6 5 5(24) 1 6 5 5(214) 1 6
5 24 5 214 5 264
Chapter 12, continued
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697Algebra 2
Worked-Out Solution Key
26. g(x) 5 24x 1 7, x0 5 1
x1 5 g(x0) x2 5 g(x1) x3 5 g(x2)
5 g(1) 5 g(3) 5 g(25)
5 24(1) 1 7 5 24(3) 1 7 5 24(25) 1 7
5 3 5 25 5 27
27. f (x) 5 1 } 2 x 2 3, x0 5 2
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (2) 5 f (22) 5 f (24)
5 1 } 2 (2) 2 3 5
1 } 2 (22) 2 3 5
1 } 2 (24) 2 3
5 22 5 24 5 25
28. f (x) 5 2 } 3 x 1 5, x0 5 6
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (6) 5 f (9) 5 f (11)
5 2 } 3 (6) 1 5 5
2 } 3 (9) 1 5 5
2 } 3 (11) 1 5
5 9 5 11 5 12 1 }
3
29. h(x) 5 x2 2 4, x0 5 23
x1 5 h(x0) x2 5 h(x1) x3 5 h(x2)
5 h(23) 5 h(5) 5 h(21)
5 (23)2 2 4 5 (5)2 2 4 5 (21)2 2 4
5 5 5 21 5 437
30. f (x) 5 2x2 1 1, x0 5 21
x1 5 f (x0) x2 5 f(x1) x3 5 f (x2)
5 f (21) 5 f (3) 5 f (19)
5 2(21)2 1 1 5 2(3)2 1 1 5 2(19)2 1 1
5 3 5 19 5 723
31. f (x) 5 x2 2 x 1 2, x0 5 1
x1 5 f (x0) x2 5 f(x1) x3 5 f (x2)
5 f (1) 5 f (2) 5 f (4)
5 12 2 1 1 2 5 22 2 2 1 2 5 42 2 4 1 2
5 2 5 4 5 14
32. g(x) 5 23x2 1 2x, x0 5 2
x1 5 g(x0) x2 5 g(x1)
5 g(2) 5 g(28)
5 23(2)2 1 2(2) 5 23(28)2 1 2(28)
5 28 5 2208
x3 5 g(x2)
5 g(2208)
5 23(2208)2 1 2(2208)
5 2130,208
33. C;
f (x) 5 22x2 1 3, x0 5 2
x1 5 f (x0) x2 5 f(x1) x3 5 f (x2)
5 f (2) 5 f (21) 5 f (5)
5 22(2) 1 3 5 22(21)2 1 3 5 22(5) 1 3
5 21 5 5 5 27
34. a1 5 3, a2 5 8
a3 5 17 5 a12 1 a2
a4 5 81 5 a22 1 a1
a5 5 370 5 a32 1 a4
a1 5 3, a2 5 8, an 5 an 2 22 1 an 2 1
35. a1 5 1, a2 5 2
a3 5 12 5 4(a1 1 a2)
a4 5 56 5 4(a2 1 a3)
a5 5 272 5 4(a3 1 a4)
a1 5 1, a2 5 2, an 5 4(an 2 2 1 an 2 1)
36. a1 5 5
a2 5 5 Ï}
3 5 Ï}
3 a1
a3 5 15 5 Ï}
3 a2
a4 5 15 Ï}
3 5 Ï}
3 a3
a5 5 45 5 Ï}
3 a4
a1 5 5, an 5 Ï}
3 an 2 1
37. a1 5 2, a2 5 5
a3 5 11 5 a2 1 3a1
a4 5 26 5 a3 1 3a2
a5 5 59 5 a4 1 3a3
a1 5 2, a2 5 5, an 5 an 2 1 1 3an 2 2
38. a1 5 8, a2 5 4
a3 5 2 5 a1
} a2
a4 5 2 5 a2
} a3
a5 5 1 5 a3
} a4
a1 5 8, a2 5 4, an 5 an 2 2
} an 2 1
39. a1 5 23, a2 5 22
a3 5 5 5 2(a2 1 a1)
a4 5 23 5 2(a3 1 a2)
a5 5 22 5 2(a4 1 a3)
a1 5 3, a2 5 22, an 5 2(an 2 1 1 an 2 2)
40. Sample answer:
a1 5 1, a2 5 2, a3 5 3, an 5 an 2 3 1 an 2 2 1 an 2 1
First eight terms: 1, 2, 3, 6, 11, 20, 37, 68
41. Sample answer:
Because x1 5 2 and x2 5 f (x1) 5 f (2) 5 2, then x3 5 f (x2) 5 f (2) 5 2, x4 5 f (x3) 5 f (2) 5 2, and so on. So, if there is a function f and an initial value x0 such that the fi rst two iterates are equal, then all of the iterates must be equal.
Chapter 12, continued
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698Algebra 2Worked-Out Solution Key
42. a. a1 5 5
a2 5 3a1 1 3 5 3(5) 1 3 5 18
a3 5 a2
} 2 5 18
} 2 5 9
a4 5 3a3 1 3 5 3(9) 1 3 5 30
a5 5 a4
} 2 5 30
} 2 5 15
a6 5 3a5 1 3 5 3(15) 1 3 5 48
a7 5 a6
} 2 5 48
} 2 5 24
a8 5 a7
} 2 5 24
} 2 5 12
a9 5 a8
} 2 5 12
} 2 5 6
a10 5 a9
} 2 5 6 } 2 5 3
b. Sample answer:
If a1 5 1, the fi rst ten terms are 1, 6, 3, 12, 6, 3, 12, 6, 3, 12.
If a1 5 2, the fi rst ten terms are 2, 1, 6, 3, 12, 6, 3, 12, 6, 3.
If a1 5 10, the fi rst ten terms are 10, 5, 18, 9, 30, 15, 48, 24, 12, 6.
The terms of the sequence will always eventually repeat the numbers 3, 12, 6.
Problem Solving
43. a. Fish at start of year n 5 0.8 p
Fish at start of year n 2 1
1 New fi sh added
an 5 0.8an 2 1 1 500
a1 5 5000, an 5 0.8an 2 1 1 500
The fi rst fi ve terms are 5000, 4500, 4100, 3780, 3524.
At the beginning of the fi fth year, there will be 3524 fi sh in the lake.
b. Over time, the population approaches 2500 fi sh.
44.
Amount of chlorine at start of week n
5 0.6 p
Amount of chlorine at start of week n 2 1
1 New chlorine added
an 5 0.6 p an 2 1 1 16
a1 5 34, an 5 0.6an 2 1 1 16
Over time, the amount of chlorine in the pool approaches 40 ounces.
45. Current balance 5 1.014 p
Previous balance
2 Payment
an 5 1.014 p an 2 1 2 100
a1 5 2000, an 5 1.014an 2 1 2 100
It will take Gladys 24 months to pay off her credit card bill. Because a24 5 62.14, the balance at the beginning of the 24th month is $62.14. So, she will be able to pay off the balance at the end of the 24th month.
46. f1 5
1 }
Ï}
5 1 1 1 Ï
} 5 }
2 2 1 2
1 }
Ï}
5 1 1 2 Ï
} 5 }
2 2 1 5 1
f2 5 1 }
Ï}
5 1 1 1 Ï
} 5 }
2 2 2 2
1 }
Ï}
5 1 1 2 Ï
} 5 }
2 2 2 5 1
f3 5 1 }
Ï}
5 1 1 1 Ï
} 5 }
2 2 3 2
1 }
Ï}
5 1 1 2 Ï
} 5 }
2 2 3 5 2
f4 5 1 }
Ï}
5 1 1 1 Ï
} 5 }
2 2 4 2
1 }
Ï}
5 1 1 2 Ï
} 5 }
2 2 4 5 3
f5 5 1 }
Ï}
5 1 1 1 Ï
} 5 }
2 2 5 2
1 }
Ï}
5 1 1 2 Ï
} 5 }
2 2 5 5 5
47. a. Current amount 5 0.70 p
Previous amount
1 New dose
an 5 0.70 p an 2 1 1 20
a1 5 20, an 5 0.70an 2 1 1 20
b. The maintenance level of the drug is 66 2 }
3 milligrams.
c. The new recursive rule would be
a1 5 2(20), an 5 0.7an 2 1 1 2(20), or
a1 5 40, an 5 0.7an 2 1 1 40.
The new maintenance level would be doubled as well,
to 133 1 }
3 milligrams.
48. a. Current balance 5 1.08 p
Previous balance
2 Amount withdrawn
an 5 1.08 p an 2 1 2 30,000
b. an 5 1.08an 2 1 2 30,000
an 1 30,000 5 1.08an 2 1
an 1 30,000
} 1.08
5 an 2 1
If a20 5 0, then a19 5 0 1 30,000
} 1.08
a18 5 a19 1 30,000
} 1.08
a0 ø 294,544.42 (use calculator).
You should have at least $294,544.42 in your account when you retire.
Mixed Review
49. a2 1 b2 5 c2 50. a2 1 b2 5 c2
32 1 32 5 x2 52 1 x2 5 92
Ï}
18 5 x x2 5 Ï}
56
3 Ï}
2 5 x x 5 2 Ï}
14
51. a2 1 b2 5 c2
122 1 x2 5 152
x2 5 Ï}
81
x 5 9
52. 163/2 5 1 2 Ï}
16 2 3 5 43 5 64
53. (2243)2/5 5 1 5 Ï}
2243 2 2 5 (23)2 5 9
54. 6421/2 5 1 }
641/2 5 1 }
2 Ï}
64 5
1 } 8
Chapter 12, continued
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699Algebra 2
Worked-Out Solution Key
Quiz 12.4–12.5 (p. 833)
1. a1 5 2, r 5 3 } 7 2. a1 5 4, r 5 2
5 } 6
S 5 a1 } 1 2 r S 5
a1 } 1 2 r
S 5 2 }
1 2 3 } 7 5
7 } 2 S 5
4 }
1 2 1 2 5 } 6 2
5 24
} 11
3. r 5 15
} 8 }
3 }
4 5
5 } 2
Because 5 }
2 ≥ 1, the series has no sum.
4. 0.777. . . 5 7(0.1) 1 7(0.1)2 1 7(0.1)3 1 . . .
5 a1 } 1 2 r
5 7(0.1)
} 1 2 0.1
5 0.7
} 0.9
5 7 } 9
5. 0.393939. . . 5 39(0.01) 1 39(0.01)2 1 39(0.01)3 1 . . .
5 a1 } 1 2 r
5 39(0.01)
} 1 2 0.01
5 0.39
} 0.99
5 13
} 33
6. 123.123123. . .
5 123 1 123(0.001) 1 123(0.001)2 1 . . .
5 a1 } 1 2 r
5 123 } 1 2 0.001
5 123
} 0.999
5 41,000
} 333
7. a1 5 2
a2 5 a1 1 4 5 2 1 4 5 6
a3 5 a2 1 4 5 6 1 4 5 10
a4 5 a3 1 4 5 10 1 4 = 14
a5 5 a4 1 4 5 14 1 4 5 18
8. a0 5 3
a1 5 (a0)2 2 5 5 32 2 5 5 4
a2 5 (a1)2 2 5 5 42 2 5 5 11
a3 5 (a2)2 2 5 5 112 2 5 = 116
a4 5 (a3)2 2 5 5 1162 2 5 5 13,451
9. a1 5 1
a2 5 4
a3 5 a2 2 a1 5 4 2 1 5 3
a4 5 a3 2 a2 5 3 2 4 5 21
a5 5 a4 2 a3 5 21 2 3 5 24
10. a1 5 5
a2 5 17
} 4 5 a1 2
3 } 4
a3 5 7 } 2 5 a2 2
3 } 4
a4 5 11
} 4 5 a3 2 3 } 4
a5 5 2 5 a4 2 3 } 4
a1 5 5, an 5 an 2 1 2 3 } 4
11. a1 5 2, a2 5 6
a3 5 12 5 a1 p a2
a4 5 72 5 a2 p a3
a5 5 864 5 a3 p a4
a1 5 2, a2 5 6, an 5 an 2 2 p an 2 1
12. a1 5 8
a2 5 24 5 3a1
a3 5 72 5 3a2
a4 5 216 5 3a3
a5 5 648 5 3a4
a1 5 8, an 5 3an 2 1
13. f (x) 5 23x 2 2, x0 5 1
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (1) 5 f (25) 5 f (13)
5 23(1) 2 2 5 23(25) 2 2 5 23(13) 2 2
5 25 5 13 5 241
14. g(x) 5 4x 1 1, x0 5 2
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (2) 5 f (9) 5 f (37)
5 4(2) 1 1 5 4(9) 1 1 5 4(37) 1 1
5 9 5 37 5 149
15. f (x) 5 22x 1 3, x0 5 22
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (22) 5 f (7) 5 f (211)
5 22(22) 1 3 5 22(7) 1 3 5 22(211) 1 3
5 7 5 211 5 25
16. f (x) 5 5x 2 7, x0 5 23
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (23) 5 f (222) 5 f (2117)
5 5(23) 2 7 5 5(222) 2 7 5 5(2117) 2 7
5 222 5 2117 5 2592
Chapter 12, continued
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700Algebra 2Worked-Out Solution Key
17. h(x) 5 x2 2 6, x0 5 21
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (21) 5 f (25) 5 f (19)
5 (21)2 2 6 5 (25)2 2 6 5 (19)2 2 6
5 25 5 19 5 355
18. f (x) 5 3x2 1 2, x0 5 0
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (0) 5 f (2) 5 f (14)
5 3(0)2 1 2 5 3(2)2 1 2 5 3(14)2 1 2
5 2 5 14 5 590
19. a1 5 25, r 5 0.85
S 5 a1 } 1 2 r
S 5 25 } 1 2 0.85 5 166
2 }
3
The pendulum swings 166 2 }
3 inches.
Problem Solving Workshop 12.5 (p. 835)
1. Let L be the limit of the sequence.
L 5 0.25L 1 300
0.75L 5 300
L 5 400
The sequence approaches 400.
2. Let L be the limit of the sequence.
L 5 0.38L 1 512
0.62L 5 512
L ø 825.8
The sequence approaches 825.8.
3. Graphing method:
Y= : nMin 5 1
u(n) 5 0.92u(n 2 1) 1 1200
u(n 2 Min) 5 50,000
WINDOW :
nMin 5 1 Xmin 5 0 Ymin 5 5000
nMax 5 1 Xmax 5 200 Ymax 5 35,000
PlotStart 5 1 Xscl 5 50 Yscl 5 5000
PlotStep 5 1
n=150X=150 Y=15000.141
Algebraic method:
a1 5 50,000, an 5 0.92an 2 1 1 1200
L 5 0.92L 1 1200
0.08L 5 1200
L 5 15,000
With each method, the number of members approaches 15,000.
4. Graphing method:
Y= : nMin 5 1
u(n) 5 0.98u(n 2 1) 1 1150
u(n 2 Min) 5 54,000
WINDOW :
nMin 5 1 Xmin 5 0 Ymin 5 45,000
nMax 5 600 Xmax 5 600 Ymax 5 60,000
PlotStart 5 1 Xscl 5 50 Yscl 5 5000
PlotStep 5 10
n=550X=550 Y=57499.947
Algebraic method:
a1 5 54,000, an 5 0.98an 2 1 1 1150
L 5 0.98L 1 1150
0.02L 5 1150
L 5 57,500
With each method, the number of books in the library approaches 57,500.
5. Each year, 2% of the books are lost or discarded, so 98% of the books remain. The coeffi cient of an 2 1 should be 0.98, not 0.02.
a1 5 54,000, an 5 0.98an 2 1 1 1150
Let L be the limit of the sequence. Then:
L 5 0.98L 1 1150
0.02L 5 1150
L 5 57,500
6. Sample answer: Save a penny on day one, and increase each day’s savings by a penny.
a1 5 0.01, an 5 an 2 1 1 0.01
12.5 Extension (p. 837)
1. Prove: i 5 1
∑ n
(2i 2 1) 5 n2
Basis Step: 2(1) 2 1 0 12 → 1 5 1 ✓
Inductive Step:
1 1 3 1 5 1 . . . 1 (2k 2 1) 5 k2
1 1 3 1 5 1 . . . 1 (2k 2 1) 1 (2(k 1 1) 2 1
5 k2 1 (2(k 1 1) 2 1)
5 k2 1 2k 1 1
5 (k 1 1)2
Therefore, i 5 1
∑ n
(2i 2 1) 5 n2 for all positive integers n.
2. Prove: i 5 1
∑ n
i2 5 n(n 1 1)(2n 1 1)
}} 6
Basis Step: 12 0 1(1 1 1)(2(1) 1 1)
}} 6 → 1 5 1 ✓
Chapter 12, continued
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701Algebra 2
Worked-Out Solution Key
Inductive Step:
1 1 4 1 9 1 . . . 1 k2 5 k(k 1 1)(2k 1 1)
}} 6
1 1 4 1 9 1 . . . 1 k2 1 (k 1 1)2
5 k(k 1 1)(2k 1 1)
}} 6 1 (k 1 1)2
5 k(k 1 1)(2k 1 1) 1 6(k 1 1)2
}}} 6
5 (k 1 1)[k(2k 1 1) 1 6(k 1 1)]
}}} 6
5 (k 1 1)(2k2 1 7k 1 6)
}} 6
5 (k 1 1)(k 1 2)(2k 1 3)
}} 6
5 (k 1 1)((k 1 1) 1 1)((2k 1 2) 1 1)
}}} 6
Therefore, i 5 1
∑ n
i2 5 n(n 1 1)(2n 1 1)
}} 6 for all positive
integers n.
3. Prove: i 5 1
∑ n
2i 2 1 5 2n 2 1
Basis Step: 21 2 1 0 21 2 1 → 1 5 1 ✓
Inductive Step:
1 1 2 1 4 1 . . . 1 2k 2 1 5 2k 2 1
1 1 2 1 4 1 . . . 1 2k 2 1 1 2(k 1 1) 2 1
5 2k 2 1 1 2(k 1 1) 2 1
5 2(2k) 2 1
5 2k 1 1 2 1
Therefore, i 5 1
∑ n
2i 2 1 5 2n 2 1 for all positive integers n.
4. Prove: i 5 1
∑ n
a1r i 2 1 5 a1 1 1 2 rn }
1 2 r 2
Basis Step: a1r 1 2 1 0 a1 1 1 2 r1 }
1 2 r 2 → a1 5 a1 ✓
Inductive Step:
a1 1 a1r 1 a1r2 1 . . . 1 a1rk 2 1 5 a1 1 1 2 r k }
1 2 r 2
a1 1 a1r 1 a1r2 1 . . . 1 a1rk 2 1 1 a1r (k 1 1) 2 1
5 a1 1 1 2 r k }
1 2 r 2 1 a1r(k 1 1) 2 1
5 a1 1 1 2 rk }
1 2 r 1 rk 2
5 a1 1 1 2 rk 1 1 }
1 2 r 2
Therefore,
i 5 1 ∑
n
a1r i2 1 5 a1 1 1 2 rn }
1 2 r 2 for all positive
integers n.
5. Prove: i 5 1
∑ n
1 }
i(i 1 1) 5
n } n 1 1
Basis Step: 1 }
1(1 1 1) 0
1 }
1 1 1 →
1 }
2 5
1 } 2 ✓
Inductive Step:
1 }
2 1
1 } 6 1
1 } 12 1 . . . 1
1 }
k(k 1 1) 5
k }
k 1 1
1 }
2 1
1 } 6 1
1 } 12 1 . . . 1
1 }
k(k 1 1) 1
1 }}
(k 1 1)(k 1 2)
5 k }
k 1 1 1
1 }}
(k 1 1)(k 1 2)
5 k(k 1 2) 1 1
}} (k 1 1)(k 1 2)
5 (k 1 1)2
}} (k 1 1)(k 1 2)
5 k 1 1
} k 1 2
5 k 1 1 }
(k 1 1) 1 1
Therefore i 5 1
∑ n
1 }
i(i 1 1) 5
n } n 1 1 for all positive integers n.
6. Prove: i 5 1
∑ n
(2i)2 5 2n(n 1 1)(2n 1 1)
}} 3
Basis Step: (2 p 1)2 0 2 p 1(1 1 1)(2 p 1 1 1)
}} 3 → 4 5 4 ✓
Inductive Step:
4 1 16 1 36 1 . . . 1 (2k)2 5 2k(k 1 1)(2k 1 1)
}} 3
4 1 16 1 36 1 . . . 1 (2k)2 1 (2(k 1 1))2
5 2k(k 1 1)(2k 1 1)
}} 3 1 (2(k 1 1))2
5 2k(k 1 1)(2k 1 1)
}} 3 1 4(k 1 1)2
5 2k(k 1 1)(2k 1 1) 1 12(k 1 1)2
}}} 3
5 2(k 1 1)[k(2k 1 1) 1 6(k 1 1)]
}}} 3
5 2(k 1 1)(k 1 2)(2k 1 3)
}} 3
5 2(k 1 1)[(k 1 1) 1 1][2(k 1 1) 1 1]
}}} 3
Therefore, i 5 1
∑ n
(2i)2 5 2n(n 1 1)(2n 1 1)
}} 3 for all positive
integers n.
7. Series: 1, 6, 15, 28. . .
A recursive formula for the nth hexagonal number is Hn 5 Hn 2 1 1 4n 2 3.
Prove: i 5 1
∑ n
4i 2 3 5 n(2n 2 1)
Basis Step: 4 p 1 2 3 0 1(2 p 1 2 1) → 1 5 1 ✓
Chapter 12, continued
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702Algebra 2Worked-Out Solution Key
Inductive Step:
1 1 5 1 9 1 . . . 1 4k 2 3 5 k(2k 2 1)
1 1 5 1 9 1 . . . 1 4k 2 3 1 4(k 1 1) 2 3
5 k(2k 2 1) 1 4(k 1 1) 2 3
5 2k2 1 3k 1 1
5 (k 1 1)(2k 1 1)
5 (k 1 1)[2(k 1 1) 2 1]
Therefore, the nth hexagonal number is given by n(2n 2 1) for all positive integers n.
8. Prove: f1 1 f2 1 f3 1 . . . 1 fn 5 fn 1 2 2 1
Basis Step: f1 0 f3 2 1 → 1 0 2 2 1 → 1 5 1 ✓
Inductive Step:
f1 1 f2 1 . . . 1 fk 5 fk 1 2 2 1
f1 1 f2 1 . . . 1 fk 1 fk 1 1 5 fk 1 2 1 fk 1 1 2 1
5 fk 1 3 2 1
5 f(k 1 1) 1 2 2 1
Therefore, the sum of the first n Fibonacci numbers is fn 1 2 2 1 for all postive integers n.
Mixed Review of Problem Solving (p. 838)
1. a. a1 5 2(12(0.7)) 5 16.8, r 5 0.7
an 5 a1rn 2 1
an 5 16.8(0.7)n 2 1
Distance 5 i 5 1
∑ `
16.8(0.7)i 2 1
b. S 5 a1 } 1 2 r
S 5 16.8
} 1 2 0.7 5 56
The ball travels a total of 56 1 12 5 68 feet.
2. a. Number of new branches: 1, 2, 4, 8, 16, 32
b. The sequence is geometric with a common ratio of 2.
c. Explicit rule: a1 5 1, r 5 2 } 1 5 2
an 5 a1rn 2 1
an 5 2n 2 1
Recursive rule: a1 5 1, r 5 2
an 5 ran 2 1
a1 5 1, an 5 2an 2 1
3. f (x) 5 x2 2 8, x0 5 2
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (2) 5 f (24) 5 f (8)
5 22 2 8 5 (24)2 2 8 5 82 2 8
5 24 5 8 5 56
x1 1 x2 1 x3 5 24 1 8 1 56 5 60
4. Sample answer: Let a1 5 4 and d 5 3.
Explicit rule: an 5 a1 1 (n 2 1)d
an 5 4 1 (n 2 1)(3) 5 1 1 3n
Recursive rule: an 5 an 2 1 1 d
a1 5 4, an 5 an 2 1 1 3
5. Sample answer:
If 21 < r < 1, then rn gets closer and closer to zero as n increases, and the sum of the series approaches the
value a1 }
1 2 r .
Sn 5 a1 1 1 2 rn }
1 2 r 2 ø a1 1 1 2 0
} 1 2 r
2 5 a1 } 1 2 r
If r < 21 or r > 1, then rn gets further away from zero as n increases, and the sum does not approach a certain number.
If r 5 1, the sum does not exist because the value is undefi ned.
Sn 5 a1 1 1 2 rn }
1 2 r 2 5 a1 1 1 2 1n
} 1 2 1
2 5 a1 1 0 } 0 2 Undefi ned
6. The length would be fi nite because the common ratio is 0.9, which is less than 1.
a1 5 16, r 5 0.9
S 5 a1 } 1 2 r
S 5 16 } 1 2 0.9 5 160
The length of the spring would be 160 inches.
7. a. 6.5
} 12
ø 0.542
The monthly interest rate is approximately 0.542%.
Amount owed 5
(1 1 interest)(Current balance) 2 Payment
Recursive rule: a1 5 10,000, an 5 1.00542an 2 1 2 196
b. After 12 months, you will owe about $8246.37.
c. a1 5 10,000; an 5 1.00542an 2 1 2 246
If you pay $246 a month, the loan will be repaid in 47 months.
d. Sample answer:
Yes, it is benefi cial to pay the extra $50 each month. This causes the balance to decrease faster, so you don’t have to pay as much interest.
8. Number of trees
5 0.9 p Previous number of trees
1 New trees
an 5 0.9an 2 1 1 500
a1 5 8000, an 5 0.9an 2 1 1 500
Over an extended period of time, there will be about 5000 trees on the farm.
Chapter 12, continued
n2ws-1200-b.indd 702 6/27/06 11:36:30 AM
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703Algebra 2
Worked-Out Solution Key
9. Sample answer:
S 5 a1 } 1 2 r
Let S 5 4 and choose r 5 1 }
2 .
4 5 a1 }
1 2 1 } 2
2 5 a1
So, i 5 1
∑ `
2 1 1 } 2 2 i 2 1
is an infi nite geometric series with a
sum of 4.
Chapter 12 Review (pp. 840–842)
1. The values in the range of a sequence are called the terms of the sequence.
2. A sequence is arithmetic if the difference between consecutive terms is constant.
3. An explicit rule gives an as a function of the term’s position number n in the sequence.
4. In a geometric sequence, the ratio of any term to the previous term is constant.
5. n 5 1
∑ 6
(n2 1 7) 5 (12 1 7) 1 (22 1 7) 1 (32 1 7) 1 (42 1 7) 1 (52 1 7) 1 (62 1 7) 5 8 1 11 1 16 1 23 1 32 1 43
5 133
6. i 5 2
∑ 6
(10 2 4i) 5 (10 2 4 p 2) 1 (10 2 4 p 3)
1 (10 2 4 p 4) 1 (10 2 4 p 5)
1 (10 2 4 p 6)
5 2 1 (22) 1 (26) 1 (210) 1 (214)
5 230
7. i 5 1
∑ 17
i 5 n(n 1 1)
} 2 (special formula)
5 17(17 1 1)
} 2
5 153
8. k 5 1
∑ 25
k2 5 n(n 1 1)(2n 1 1)
}} 6 (special formula)
5 25(25 1 1)(2 p 25 1 1)
}} 6
5 5525
9. a1 5 8, d 5 5 2 8 5 23
an 5 a1 1 (n 2 1)d
an 5 8 1 (n 2 1)(23)
an 5 11 2 3n
10. a8 5 54, d 5 7
an 5 a1 1 (n 2 1)d
a8 5 a1 1 (8 2 1)d
54 5 a1 1 7(7)
5 5 a1
an 5 5 1 (n 2 1)(7)
an 5 22 1 7n
11. a4 5 27, a11 5 69
a11 5 a1 1 (11 2 1)d → 69 5 a1 1 10d
a4 5 a1 1 (4 2 1)d → 27 5 a1 1 3d
42 5 7d
6 5 d
69 5 a1 1 10(6) → a1 5 9
an 5 a1 1 (n 2 1)d
an 5 9 1 (n 2 1)(6)
an 5 3 1 6n
12. a1 5 3 1 2(1) 5 5, a15 5 3 1 2(15) 5 33
Sn 5 n 1 a1 1 an }
2 2
S15 5 15 1 a1 1 a15 }
2 2
S15 5 15 1 5 1 33 }
2 2 5 285
13. a1 5 25 2 3(1) 5 22, a26 5 25 2 3(26) 5 253
Sn 5 n 1 a1 1 an }
2 2
S26 5 26 1 a1 1 a26 }
2 2
S26 5 26 1 22 1 (253) }
2 2 5 2403
14. a1 5 6(1) 2 5 5 1, a22 5 6(22) 2 5 5 127
Sn 5 n 1 a1 1 an }
2 2
S22 5 22 1 a1 1 a22 }
2 2
S22 5 22 1 1 1 127 }
2 2 5 1408
15. a1 5 284 1 8(1) 5 276, a30 5 284 1 8(30) 5 156
Sn 5 n 1 a1 1 an }
2 2
S30 5 30 1 a1 1 a30 }
2 2
S30 5 30 1 276 1 156 }
2 2 5 1200
Chapter 12, continued
n2ws-1200-b.indd 703 6/27/06 11:36:34 AM
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704Algebra 2Worked-Out Solution Key
16. a1 5 200 1 25 5 225, d 5 25
an 5 a1 1 (n 2 1)d
an 5 225 1 (n 2 1)(25)
an 5 200 1 25n
After n months, Joe will have paid 200 1 25n dollars toward the computer.
17. a1 5 256, r 5 64
} 256 5 1 } 4
an 5 a1rn 2 1
an 5 256 1 1 } 4 2 n 2 1
18. a2 5 200, r 5 5
an 5 a1rn 2 1
a2 5 a1r2 2 1
200 5 a1(5)
40 5 a1
an 5 40(5)n 2 1
19. a1 5 144, a3 5 16
an 5 a1rn 2 1
a3 5 a1r3 2 1
16 5 144r2
1 }
9 5 r2
6 1 } 3 5 r
an 5 144 1 1 } 3 2
n 2 1 or an 5 144 1 2
1 } 3 2 n 2 1
20. a1 5 3, r 5 5
S6 5 a1 1 1 2 r6 }
1 2 r 2
S6 5 3 1 1 2 56 }
1 2 5 2 5 11,718
21. a1 5 8, r 5 2
S9 5 a1 1 1 2 r9 }
1 2 r 2
S9 5 8 1 1 2 29 }
1 2 2 2 5 4088
22. a1 5 15, r 5 2 } 3
S5 5 a1 1 1 2 r5 }
1 2 r 2
S5 5 15 1 1 2 1 2 } 3 2 5 }
1 2 2 } 3 2 5
1055 } 27
23. a1 5 40, r 5 1 } 2
S7 5 a1 1 1 2 r7 }
1 2 r 2
S7 5 40 1 1 2 1 1 } 2 2 7 }
1 2 1 } 2 2 5
635 } 8
24. a1 5 3, r 5 5 } 8
S 5 a1 } 1 2 r
S 5 3 }
1 2 5 } 8 5 8
25. a1 5 7, r 5 2 3 } 4
S 5 a1 } 1 2 r
S 5 7 }
1 2 1 2 3 } 4 2 5 4
26. r 5 1.3
Because r ≥ 1, the series has no sum.
27. a1 5 20.2, r 5 0.5
S 5 a1 } 1 2 r
S 5 20.2
} 1 2 0.5 5 0.4
28. 0.888. . . 5 8(0.1) 1 8(0.1)2 1 8(0.1)3. . .
5 a1 } 1 2 r
5 8(0.1)
} 1 2 0.1
5 0.8
} 0.9
5 8 } 9
29. 0.546546546. . . 5
546(0.001) 1 546(0.001)2 1 546(0.001)3. . .
5 a1 } 1 2 r
5 546(0.001)
} 1 2 0.001
5 0.546
} 0.999
5 182
} 333
30. 0.3787878. . . 5 0.37 1 87(0.01)2 1 87(0.01)3 1 . . .
5 37
} 100 1 a1 } 1 2 r
5 37
} 100 1 87(0.01)2
} 1 2 0.01
5 37
} 100 1 0.0087
} 0.99
5 37
} 100 1 87 } 9900
5 25
} 66
Chapter 12, continued
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705Algebra 2
Worked-Out Solution Key
31. 0.783838. . . 5 0.78 1 38(0.01)2 1 38(0.01)3 1 . . .
5 78
} 100 1 a1 } 1 2 r
5 78
} 100 1 38(0.01)2
} 1 2 0.01
5 78
} 100 1 0.0038
} 0.99
5 78
} 100 1 38 } 9900
5 388
} 495
32. a1 5 4
a2 5 a1 1 9 5 4 1 9 5 13
a3 5 a2 1 9 5 13 1 9 5 22
a4 5 a3 1 9 5 22 1 9 5 31
a5 5 a4 1 9 5 31 1 9 5 40
33. a1 5 8
a2 5 5a1 5 5(8) 5 40
a3 5 5a2 5 5(40) 5 200
a4 5 5a3 5 5(200) 5 1000
a5 5 5a4 5 5(1000) 5 5000
34. a1 5 2
a2 5 2a1 5 2(2) 5 4
a3 5 3a2 5 3(4) 5 12
a4 5 4a3 5 4(12) 5 48
a5 5 5a4 5 5(48) 5 240
35. a1 5 6, r 5 18
} 6 5 3
an 5 ran 2 1
a1 5 6, an 5 3an 2 1
36. The nth term is n greater than the previous term.
an 5 an 2 1 1 d
a1 5 4, an 5 an 2 1 1 n
37. a1 5 7, d 5 13 2 7 5 6
an 5 an 2 1 1 d
a1 5 7, an 5 an 2 1 1 6
38. Current population
5 1.01 p Previous population
an 5 1.01an 2 1
a1 5 26,000, an 5 1.01an 2 1
Chapter 12 Test (p. 843)
1. a2 2 a1 5 9 2 5 5 4
a3 2 a2 5 13 2 9 5 4
a4 2 a3 5 17 2 13 5 4
Arithmetic; the common difference is 4.
2. a2
} a1 5
6 } 3 5 2
a3
} a2 5
12 } 6 5 2
a4
} a3 5
24 } 12 5 2
Geometric; the common ratio is 2.
3. a2
} a1 5
10 } 40 5
1 } 4
a3
} a2 5
5 } 2 } 10 5
1 } 4
a4
} a3 5
5 } 8 }
5 }
2 5
1 } 4
Geometric; the common ratio is 1 }
4 .
4. Neither, there is no common ratio or common difference.
5. a1 5 6 2 12 5 5 6. a1 5 7 p 13 5 7
a2 5 6 2 22 5 2 a2 5 7 p 23 5 56
a3 5 6 2 32 5 23 a3 5 7 p 33 5 189
a4 5 6 2 42 5 210 a4 5 7 p 43 5 448
a5 5 6 2 52 5 219 a5 5 7 p 53 5 875
a6 5 6 2 62 5 230 a6 5 7 p 63 5 1512
7. a1 5 4
a2 5 5a1 5 5(4) 5 20
a3 5 5a2 5 5(20) 5 100
a4 5 5a3 5 5(100) 5 500
a5 5 5a4 5 5(500) 5 2500
a6 5 5a5 5 5(2500) 5 12,500
8. a1 5 21
a2 5 a1 1 6 5 21 1 6 5 5
a3 5 a2 1 6 5 5 1 6 5 11
a4 5 a3 1 6 5 11 1 6 5 17
a5 5 a4 1 6 5 17 1 6 5 23
a6 5 a5 1 6 5 23 1 6 5 29
9. Next term: 29
a1 5 5, d 5 11 2 5 5 6
an 5 a1 1 (n 2 1)d
an 5 5 1 (n 2 1)(6) 5 21 1 6n
10. Next term: 1875
a1 5 3, r 5 15
} 3 5 5
an 5 a1r n 2 1
an 5 3(5)n 2 1
Chapter 12, continued
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706Algebra 2Worked-Out Solution Key706Algebra 2Worked-Out Solution Key
Chapter 12, continued
11. Next term: 10
} 25
a1 5 6 } 5 5
1 1 5 }
5(1)
a2 5 7 } 10 5
2 1 5 }
5(2)
a3 5 8 } 15 5
3 1 5 }
5(3)
an 5 n 1 5
} 5n
12. Next term: 8
a1 5 1.6 5 1.6(1)
a2 5 3.2 5 1.6(2)
a3 5 4.8 5 1.6(3)
an 5 1.6n
13. i 5 1
∑ 48
i 5 n(n 1 1)
} 2 5 48(48 1 1)
} 2 5 1176
14. n 5 1
∑ 28
n2 5 n(n 1 1)(2n 1 1)
}} 6 5 28(28 1 1)(2(28) 1 1)
}} 6 5 7714
15. a1 5 4(1) 2 9 5 25, a10 5 4(10) 2 9 5 31
Sn 5 n 1 a1 1 an }
2 2
S10 5 10 1 a1 1 a10 }
2 2
S10 5 10 1 25 1 31 }
2 2 5 130
16. a1 5 2(1) 1 5 5 7, a19 5 2(19) 1 5 5 43
Sn 5 n 1 a1 1 an }
2 2
S19 5 19 1 a1 1 a19 }
2 2
S19 5 19 1 7 1 43 }
2 2 5 475
17. a1 5 9, r 5 2
Sn 5 a1 1 1 2 rn }
1 2 r 2
S5 5 a1 1 1 2 r5 }
1 2 r 2
S5 5 9 1 1 2 25 }
1 2 2 2 5 279
18. a1 5 12, r 5 1 } 3
Sn 5 a1 1 1 2 rn }
1 2 r 2
S6 5 a1 1 1 2 r6 }
1 2 r 2
S6 5 12 1 1 2 1 1 } 3 2 6 }
1 2 1 1 } 3 2 2 5
1456 } 81
19. a1 5 8, r 5 3 } 4 20. a1 5 20, r 5
3 } 10
S 5 a1 } 1 2 r S 5
a1 } 1 2 r
S 5 8 }
1 2 1 3 } 4 2 5 32 S 5
20 }
1 2 1 3 } 10 2 5
200 } 7
21. 0.111. . . 5 1(0.1) 1 1(0.1)2 1 1(0.1)3 1 . . .
5 a1 } 1 2 r
5 1(0.1)
} 1 2 0.1
5 0.1
} 0.9
5 1 } 9
22. 0.464646. . . 5 46(0.01) 1 46(0.01)2 1 46(0.01)3 1 . . .
5 a1 } 1 2 r
5 46(0.01)
} 1 2 0.01
5 0.46
} 0.99
5 46
} 99
23. 0.187187187. . . 5 187(0.001) 1 187(0.001)2
1 187(0.001)3 1 . . .
5 a1 } 1 2 r
5 187(0.001)
} 1 2 0.001
5 0.187
} 0.999
5 187
} 999
24. 0.3252525. . . 5 0.32 1 52(0.01)2 1 52(0.01)3 1 . . .
5 32
} 100 1 a1 } 1 2 r
5 32
} 100 1 52(0.01)2
} 1 2 0.01
5 32
} 100 1 0.0052
} 0.99
5 32
} 100 1 52 } 9900
5 161
} 495
25. a1 5 2 26. a1 5 3
a2 5 12 5 6a1 a2 5 10 5 a1 1 7
a3 5 72 5 6a2 a3 5 17 5 a2 1 7
a4 5 432 5 6a3 a4 5 24 5 a3 1 7
a1 5 2, an 5 6an 2 1 a1 5 3, an 5 an 2 1 1 7
Chapter 12, continued
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707Algebra 2
Worked-Out Solution Key
27. a1 5 135
a2 5 45 5 1 } 3 a1
a3 5 15 5 1 } 3 a2
a4 5 5 5 1 } 3 a3
a1 5 135, an 5 1 } 3 an 2 1
28. a1 5 1
a2 5 23 5 23a1
a3 5 9 5 23a2
a4 5 227 5 23a3
a1 5 1, an 5 23an 2 1
29. f (x) 5 3x 2 7, x0 5 4
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (4) 5 f (5) 5 f (8)
5 3(4) 2 7 5 3(5) 2 7 5 3(8) 2 7
5 5 5 8 5 17
30. f (x) 5 8 2 5x, x0 5 1
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (1) 5 f (3) 5 f (27)
5 8 2 5(1) 5 8 2 5(3) 5 8 2 5(27)
5 3 5 27 5 43
31. f (x) 5 x2 1 2, x0 5 21
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (21) 5 f (3) 5 f (11)
5 (21)2 1 2 5 32 1 2 5 112 1 2
5 3 5 11 5 123
32. a. n represents the number of rows and columns.
an represents the number of blue squares.
b. n 1 2 3 4 5 6 7 8
an 1 2 5 8 13 18 25 32
c. an 5 n2
} 2 1 1 } 4 [1 2 (21)n]
a1 5 12
} 2 1 1 } 4 [1 2 (21)1] 5 1
a2 5 22
} 2 1 1 } 4 [1 2 (21)2] 5 2
a3 5 32
} 2 1 1 } 4 [1 2 (21)3] 5 5
a4 5 42
} 2 1 1 } 4 [1 2 (21)4] 5 8
a5 5 52
} 2 1 1 } 4 [1 2 (21)5] 5 13
a6 5 62
} 2 1 1 } 4 [1 2 (21)6] 5 18
a7 5 72
} 2 1 1 } 4 [1 2 (21)7] 5 25
a8 5 82
} 2 1 1 } 4 [1 2 (21)8] 5 32
These values are the same as the values in the table, so this rule defi nes the sequence represented by the checkerboard quilts.
33. a1 5 3072, r 5 1 } 4
an 5 a1rn 2 1
an 5 3072 1 1 } 4 2 n 2 1
There are a5 5 3072 1 1 } 4 2 5 2 1
5 12 actors in the
fi fth round of auditions. After this round, three actors will remain, which is exactly the number required for the three main parts in the play. So, the rule makes sense for 1 ≤ n ≤ 5.
Standardized Test Preparation (p. 845)
1. B;
an 5 2n 2 5
a15 5 2(15) 2 5 5 25
2. A;
Sn 5 n 1 a1 1 an }
2 2
S14 5 14 1 a1 1 a14 }
2 2
S14 5 14 1 23 1 23 }
2 2 5 140
Standardized Test Practice (pp. 846–847)
1. C;
S 5 a1 } 1 2 r
Basketball: a1 5 7.2, r 5 0.36
S 5 7.2 } 1 2 0.36 5 11.25
Baseball: a1 5 6, r 5 0.30
S 5 6 } 1 2 0.3 ø 8.57
Difference 5 11.25 2 8.57 5 2.68
2. C;
a1 5 20, r 5 10
} 20 5 0.5
an 5 ran 2 1
a1 5 20, an 5 0.5an 2 1
3. D;
The sequence is 12, 11, 9, 6, 2, 23 . . ..
There is no common difference or common ratio, so this sequence is neither geometric nor arithmetic.
4. B;
a1 5 1 5 20
a2 5 2 5 21
a3 5 4 5 22
a4 5 8 5 23
an 5 2n 2 1
Chapter 12, continued
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708Algebra 2Worked-Out Solution Key
5. D;
a1 5 1, r 5 2
Sn 5 a1 1 1 2 rn }
1 2 r 2
S8 5 a1 1 1 2 r8 }
1 2 r 2
S8 5 1 1 1 2 28 }
1 2 2 2 5 255
6. B;
a1 5 2 5 1 p 2
a2 5 6 5 2 p 3
a3 5 12 5 3 p 4
a4 5 20 5 4 p 5
an 5 n(n 1 1)
7. D;
n 5 4, an 5 n(n 1 1)
Sum 5 i 5 1
∑ 4
i(i 1 1)
8. a3 5 12, a5 5 48
a3 5 a1r3 2 1 → 12 5 a1r2 → a1 5 12
} r2
a5 5 a1r5 2 1 → 48 5 a1r4
48 5 1 12 }
r2 2 r4
48 5 12r2
4 5 r2
62 5 r
a1 5 12 }
(62)2 → a1 5 3
9. The sequence is geometric.
a1 5 24, r 5 12
} 24 5 23
an 5 a1rn 2 1
an 5 24(23)n 2 1
a8 5 24(23)8 2 1 5 8748
10. i 5 1
∑ 5
0.5(2)i 2 1 5 0.5 1 1 1 2 1 4 1 8 5 15.5
11. 0.151515. . . 5 15(0.01) 1 15(0.01)2 1 15(0.01)3 1 . . .
5 a1 } 1 2 r
5 15(0.01)
} 1 2 0.01
5 0.15
} 0.99
5 15
} 99
5 5 } 33
12. f (x) 5 2x 2 1, x0 5 2
x1 5 f (x0) x2 5 f (x1) x3 5 f (x2)
5 f (2) 5 f (3) 5 f (5)
5 2(2) 2 1 5 2(3) 2 1 5 2(5) 2 1
5 3 5 5 5 9
x1 1 x2 1 x3 5 3 1 5 1 9 5 17
13. a1 5 0.5 a5 5 2(39) 1 5 5 83
a2 5 2(0.5) 1 5 5 6 a6 5 2(83) 1 5 5 171
a3 5 2(6) 1 5 5 17 a7 5 2(171) 1 5 5 347
a4 5 2(17) 1 5 5 39 a8 5 2(347) 1 5 5 699
14. an 5 6n 1 3
a1 5 6(1) 1 3 5 9, a15 5 6(15) 1 3 5 93
Sn 5 n 1 a1 1 an }
2 2
S15 5 15 1 9 1 93 }
2 2 5 765
15. d 5 6 2 3 5 3
16. First round 5 10 people
Second round 5 10 p 10 people
Third round 5 10 p 10 p 10 people
nth round 5 10n people
If 100 million people already received the email, then you are in the 8th round because 108 5 100 million.
17. dn 5 1 } 2 n(n 2 3), n ≥ 3
d3 5 1 } 2 (3)(3 2 3) 5 0
d4 5 1 } 2 (4)(4 2 3) 5 2
d5 5 1 } 2 (5)(5 2 3) 5 5
d6 5 1 } 2 (6)(6 2 3) 5 9
d7 5 1 } 2 (7)(7 2 3) 5 14
d8 5 1 } 2 (8)(8 2 3) 5 20
There is no common ratio or common difference between consecutive terms, so the sequence is neither geometric nor arithmetic.
18. This situation should be represented by a series.
Sample answer: The company is pledging a total amount, not just a sequence of values.
19. a. The sequence is arithmetic because each radius is 1.22 meters greater than the previous radius.
b. a1 5 36.5, d 5 1.22
an 5 a1 1 (n 2 1)d
an 5 36.5 1 (n 2 1)(1.22) 5 35.28 1 1.22n
c. a8 5 35.28 1 1.22(8) 5 45.04
The track meets the requirement because the curve radius of lane 8 is 45.04 meters, which is less than the 50-meter maximum.
Chapter 12, continued
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709Algebra 2
Worked-Out Solution Key
20. a. Current balance
5 Interest p Previous balance 2 Payment
an 5 1.0075 p an 2 1 2 300
a1 5 16,000, an 5 1.0075an 2 1 2 300
b. a18 ø 12,749.33
Mark will owe $12,749.33 at the beginning of the 18th month.
c. Mark will pay the loan off in 69 months, with the last payment being about $109.00.
d. If the payments are $350, then Mark can have the loan paid off in 57 months. He will pay less overall because the balance is less each month and the balance is paid off faster, so less interest is paid.
Cumulative Review, Chs. 1–12 (pp. 848–849)
1. 3x 2 y 5 5 2. 1 }
2 x 1 3y 5 24
1
x
y
21
1
x
y
21
3. y 5 x 1 3 2 8 4. y 5 x2 2 6x 2 27
1
x
y
21
5
x
y
24
5. y 5 22(x 1 6)(x 2 1) 6. y 5 (x 2 3)2 1 4
5
x
y
21
1
x
y
21
7. y 5 Ï}
x 1 6 8. y 5 3 Ï}
x 2 2
1
x
y
21
1
x
y
21
9. y 5 3 p 4x 2 2 10. y 5 12 1 1 } 8 2
x
1
x
y
21
1
x
y
21
11. y 5 2 } x 2 3 1 5 12. y 5
6 }
x2 2 4
1
x
y
21
x
y
21
1
13. 2 3 1 8 5 2(8) 2 1(3) 5 13
14. 12 3 27 8 5 12(8) 2 (27)(3) 5 117
15. 0 5 2 10 13 24
25 4 21 0 5
10 13
25 4
5 (0 1 100 1 80) 2 (2130 1 0 2 50)
5 360
16. 5 29 4 4 2 1
0 1 1 5 29
4 2
0 1
5 (10 1 0 1 16) 2 (0 1 5 2 36)
5 57
17. y 5 a } x
6 5 a } 3 → a 5 18
y 5 18
} x
When x 5 28: y 5 18
} 28 5 2
9 } 4
18. y 5 a } x
9 5 a }
24 → a 5 236
y 5 236
} x
When x 5 28: y 5 236
} 28 5
9 } 2
19. y 5 a } x
1 }
8 5
a } 4 → a 5
1 } 2
y 5 1 } 2 } x 5
1 } 2x
When x 5 28: y 5 1 }
2(28) 5 2
1 } 16
Chapter 12, continued
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710Algebra 2Worked-Out Solution Key
20. y 5 a } x
2 } 5 5
a } 9 → a 5
18 } 5
y 5 18
} 5 } x 5
18 } 5x
When x 5 28: y 5 18 }
5(28) 5 2
9 } 20
21. x2
} 36
1 y2
} 4 5 1 4
x
y
22
x2 }
62 1 y2
} 22 5 1
Vertices: (6, 0), (26, 0)
Co-vertices: (0, 2), (0, 22)
22. y2
} 100
2 x2
} 49 5 1
x
y
22
4
y2
} 102 2
x2
} 72 5 1
Transverse axis: Vertical
Asymptotes: y 5 10
} 7 x,
y 5 2 10
} 7 x
Vertices: (0, 10), (0, 210)
23. (x 2 3)2 5 16y
2
x
y
21
(x 2 3)2 5 16(y 1 0)
Vertex: (3, 0)
24. 9P3 5 9! }
(9 2 3)! 5
9! } 6! 5 504
25. 16P5 5 16! }
(16 2 5)! 5
16! } 11! 5 524,160
26. 7C2 5 7! } 5! p 2! 5 21
27. 6C6 5 6! } 6! p 0! 5
6! } 6! 5 1
28. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.85 5 0.32 1 0.6 2 P(A and B)
P(A and B) 5 0.32 1 0.6 2 0.85 5 0.07
29. P(A and B) 5 P(A) p P(BA) P(A and B) 5 0.5 p 0.3 5 0.15
30. P(A and B) 5 P(A) p P(B)
0.2 5 0.25 p P(B)
0.8 5 P(B)
31. Mean: } x 5 19 1 11 1 8 1 10 1 11 1 15 1 16
}}} 7 ø12.857
Median: 11
Mode: 11
Range: 19 2 8 5 11
Std. Dev:
� 5 Î}}}}}
(19 2 12.857)2 1 (11 2 12.857)2 1 . . . 1 (16 2 12.857)2
}}}}} 7
ø 3.603
32. Mean: } x 5 54 1 58 1 49 1 60 1 63 1 58 1 42
}}} 7 ø 54.857
Median: 58
Mode: 58
Range: 63 2 42 5 21
Std. Dev:
� 5 Î}}}}}
(54 2 54.857)2 1 (58 2 54.857)2 1 . . . 1 (42 2 54.857)2
}}}}} 7
ø 6.685
33. Mean: } x 5 216 1 203 1 225 1 216 1 212 1 228 1 209
}}}} 7
ø 215.57
Median: 216
Mode: 216
Range: 228 2 203 5 25
Std. Dev:
5 Î}}}}}
(216 2 215.57)2 1 (203 2 215.57)2 1 . . . 1 (209 2 215.57)2
}}}}} 7
ø 8.086
34. Mean: } x 5 23 1 5 2 11 1 6 2 3 1 2
}} 6 5 2 2 } 3
Median: 23 1 2
} 2 5 2
1 } 2
Mode: 23
Range: 6 2 (211) 5 17
Std. Dev:
5 Î}}}}
1 23 2 1 2
2 } 3 2 2 2 1 1 5 2 1 2
2 } 3 2 2 2 1 . . . 1 1 2 2 1 2 } 3 2 2 2
}}}} 6
ø 5.793
35. Mean: } x 5 99 1 92 1 93 1 82 1 88 1 71 1 97
}}} 7 ø 88.857
Median: 92
Mode: none
Range: 99 2 71 5 28
Std. Dev:
� 5 Î}}}}}
(99 2 88.857)2 1 (92 2 88.857)2 1 . . . 1 (97 2 88.857)2
}}}}} 7
ø 8.967
Chapter 12, continued
n2ws-1200-b.indd 710 6/27/06 11:37:19 AM
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711Algebra 2
Worked-Out Solution Key
36. Mean: } x 5 78 1 4 1 28 1 57 1 88 1 24 1 57 1 37 1 65
}}}} 9
ø 48.667
Median: 57
Mode: 57
Range: 88 2 4 5 84
Std. Dev:
5 Î}}}}}
(78 2 48.667)2 1 (4 2 48.667)2 1 . . . 1 (65 2 48.667)2
}}}}} 9
ø 25.777
37. i 5 1
∑ 6
3i2 5 3(12) 1 3(22) 1 3(32) 1 3(42) 1 3(52) 1 3(62)
5 273
38. i 5 1
∑ 16
(22 1 i) 5 (22 1 1) 1 (22 1 2) 1 . . .
1 (22 1 15) 1 (22 1 16) 5 104
39. a1 5 1, r 5 2 } 3
Sn 5 a1 1 1 2 rn }
1 2 r 2
S12 5 1 1 1 2 1 2 } 3 2 12
} 1 2
2 } 3 2 ø 2.977
40. a1 5 5, r 5 1 } 3
S 5 a1 } 1 2 r
S 5 5 }
1 2 1 } 3 5 7.5
41. Explicit rule:
a1 5 27, d 5 23 2 (27) 5 4
an 5 a1 1 (n 2 1)d
an 5 27 1 (n 2 1)(4) 5 211 1 4n
Recursive rule:
an 5 an 2 1 1 d
a1 5 27, an 5 an 2 1 1 4
42. Explicit rule:
a1 5 1, d 5 214 2 1 5 215
an 5 a1 1 (n 2 1)d
an 5 1 1 (n 2 1)(215) 5 16 2 15n
Recursive rule:
an 5 an 2 1 1 d
a1 5 1, an 5 an 2 1 2 15
43. Explicit rule:
a1 5 3, r 5 12
} 3 5 4
an 5 a1rn 2 1
an 5 3(4)n 2 1
Recursive rule:
an 5 ran 2 1
a1 5 3, an 5 4an 2 1
44. x 5 bags of cookies
Bags of cookies
Cal
end
ars
0 300 600 900 1200 x
y
0
100
200
300 y 5 calendars
2x 1 7y 5 2400
When y 5 200:
2x 1 7(200) 5 2400
2x 5 1000
x 5 500
If you sell 200 calendars, you will need to sell 500 bags of cookies to meet your goal.
45. V 5 75π, h 5 r 1 4
V 5 πr2h
} 3
75π 5 πr 2(r 1 4)
} 3
225π 5 πr2(r 1 4)
225 5 r3 1 4r2
0 5 r3 1 4r2 2 225
Possible zeros: 61, 63, 65, 69, 615, 625, 645, 675, 6225
Test r 5 1:
1 1 4 0 2225
1 5 5
1 5 5 2220 → 1 is not a zero.
Test r 5 5:
5 1 4 0 2225
5 45 225
1 9 45 0 → 5 is a zero.
r3 1 4r2 2 225 5 (r 2 5)(r2 1 9r 1 45) Because r2 1 9r 1 45 does not have any real roots,
r 5 5 is the only zero. The radius of the cone should be 5 inches.
46. I 5 17, R 5 6.5
I 5 Î}
P
} R
17 5 Î}
P }
6.5
289 5 P } 6.5
1878.5 5 P
The hair dryer consumes 1878.5 watts of power.
Chapter 12, continued
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712Algebra 2Worked-Out Solution Key
47. a1 5 18,600(0.845) 5 15,717, r 5 0.845
an 5 a1rn 2 1
an 5 15,717(0.845)n 2 1
8000 5 15,717(0.845)n 2 1
0.509 ø 0.845n 2 1
log 0.509 ø (n 2 1) log 0.845
4 ø n 2 1
5 ø n
The car will be worth $8000 in about 5 years.
48. Let the center of the circle be (0, 0) with radius 330
} 2 ,
or r 5 165.
x2 1 y2 5 r2
x2 1 y2 5 1652
x2 1 y2 5 27,225
When y 5 115:
25
25
x
y
(0, 115)
50
(0, 165)
(20 35, 115)(220 35, 115)
x2 1 1152 5 27,225
x2 5 14,000
x 5 620 Ï}
35
Distance between (220 Ï}
35 , 115) and (20 Ï}
35 , 115):
d 5 Ï}}}}
(20 Ï}
35 2 (220 Ï}
35 ))2 1 (115 2 115)2
5 Ï}
(40 Ï}
35 )2
5 40 Ï}
35
ø 236.64
The rope is about 237 feet long.
49. Let events A, B, C, and D be choosing item 1, item 2, item 3, and item 4, respectively, so that all items are different.
P(A) p P(B) p P(C) p P(D) 5 20
} 20 p 19 }
20 p 18
} 20
p 17 }
20
5 116,280
} 160,000 ø 0.727
The probability that all four people order a different item is about 0.727, or 72.7%.
50. a. Mean:
ø $202,957.14
Median: $201,900
Mode: none
Range: $23,400
Std. Dev.:
s 5 Î}}}}}
(201,900 2 202,957.14)2 1 . . . 1 (192,100 2 202,957.14)2
}}}}} 7
ø $7519.85
b. Commissions:
$201,900(0.05) 5 $10,095
$205,200(0.05) 5 $10,260
$195,800(0.05) 5 $9790
$210,300(0.05) 5 $10,515
$199,900(0.05) 5 $9995
$215,500(0.05) 5 $10,775
$192,100(0.05) 5 $9605
Mean:
x 5 10,095 1 10,260 1 9790 1 10,515 1 9995 1 10,775 1 9605
}}}}} 7
ø $10,147.86
Median: $10,095
Mode: none
Range: $1170
Std. Dev.:
s 5 Ï}}}}
(10,095 2 10,147.86)2 1 . . . 1 (9605 2 10,147.86)2
}}}} 7
ø $375.99
c. Each statistical measure in part (b) is 5% of the corresponding statistical measure in part (a).
51. a1 5 31,000, d 5 1600
an 5 a1 1 (n 2 1)d
an 5 31,000 1 (n 2 1)(1600) 5 29,400 1 1600n
a9 5 29,400 1 1600(9) 5 43,800
In the ninth year, the accountant’s salary will be $43,800.
Chapter 12, continued
} x 5
201,900 1 205,200 1 195,800 1 210,300 1 199,900 1 215,500 1 192,100
}}}}}} 7
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