1. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
(For help, go to Skills Handbook page 838.)
Find the next two numbers of each pattern. Then write a rule to describe the pattern.
1. 1, 3, 5, 7, 9, 11, –2, –4, –6, –8, –10, –12, . . .
3. 0.2, 1, 5, 25, 125, 625, , 45, 40, 35, 30, 25, . . .
5. 512, 256, 128, 64, 32, 16, , 5, 8, 11, 14, 17, . . .
7. 16, 32, 64, –3, –7, –11, –15, . . .
11-1

2. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
Solutions
1. 1, 3, 5, 7, 9, 11, 13, 15; add 2
2. –2, –4, –6, –8, –10, –12, –14, –16; subtract 2
3. 0.2, 1, 5, 25, 125, 625, 3125, 15,625; multiply by 5
4. 50, 45, 40, 35, 30, 25, 20, 15; subtract 5
5. 512, 256, 128, 64, 32, 16, 8, 4; divide by 2
6. 2, 5, 8, 11, 14, 17, 20, 23; add 3
7. 16, 32, 64, 128, 256; multiply by 2
8. –3, –7, –11, –15, –19, –23; subtract 4
11-1

3. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
a. Start with a square with sides 1 unit long. On the right side, add on a square of the same size. Continue adding one square at a time in this way. Draw the first four figures of the pattern.
b. Write the number of 1-unit segments in each figure above as a sequence.
4, 7, 10, 13, . . .
c. Predict the next term of the sequence. Explain your choice.
Each term is 3 more than the preceding term.
The next term is , or 16.
There will be 16 segments in the next figure in the pattern.
11-1

4. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
Suppose you drop a ball from a height of 100 cm. It bounces back to 80% of its previous height. How high will it go after its fifth bounce?
Original height of ball: 100 cm
After first bounce: 80% of 100 = 0.80(100) = 80
After 2nd bounce: 0.80(80) = 64
After 3rd bounce: 0.80(64) = 51.2
After 4th bounce: 0.80(51.2) = 40.96
After 5th bounce: 0.80(40.96) =
The ball will rebound about 32.8 cm after the fifth bounce.
11-1

5. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
a. Describe the pattern that allows you to find the next term in the sequence 2, 6, 18, 54, 162, Write a recursive formula for the sequence.
Multiply a term by 3 to find the next term.
A recursive formula is an = an – 1 • 3, where a1 = 2.
b. Find the sixth and seventh terms in the sequence.
Since a5 = 162
a6 = 162 • 3 = 486
and a7 = 486 • 3 = 1458
11-1

6. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
(continued)
c. Find the value of a10 in the sequence.
The term a10 is the tenth term.
a10 = a9 • 3
= (a8 • 3) • 3
= ((a7 • 3) • 3) • 3
= ((1458 • 3) • 3) • 3
= 39,366
11-1

7. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
The spreadsheet shows the perimeters of regular pentagons with sides from 1 to 4 units long. The numbers in each row form a sequence.
A B C D E
1 a1 a2 a3 a4
2 Length of Side 1  2 3  4
3 Perimeter
a. For each sequence, find the next term (a5) and the twentieth term (a20).
In the sequence in row 2, each term is the same as its subscript. Therefore, a5 = 5 and a20 = 20.
In the sequence in row 3, each term is 5 times its subscript. Therefore, a5 = 5(5) = 25 and a20 = 5(20) = 100.
11-1

8. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
(continued)
A B C D E
1 a1 a2 a3 a4
2 Length of Side 1  2 3  4
3 Perimeter
b. Write an explicit formula for each sequence.
The explicit formula for the sequence in row 2 is an = n. The explicit formula for the sequence in row 3 is an = 5n.
11-1

9. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
pages 591–593  Exercises
1. Subtract 3; 65, 62, 59.
2. Multiply by 2; 128, 256, 512.
3. Add one more to each term
(add 3, add 4, add 5, etc.); 25, 33, 42.
4. Add 3; 16, 19, 22.
5. Divide by 10; 0.001, ,
6. Multiply by ; , ,
7. Multiply by –2; –128, 256, –512.
8. Each term is the preceding term
multiplied by n; 720, 5040, 40,320.
9. Every odd-numbered term is 0, and
every even-numbered term is ; 0, , 0.
10.
11.
12. an = an–1 + 1, a1 = –2; 3
13. an = an–1 – 2, a1 = 43; 33
14. an = , a1 = 40;
15. an = an–1 – 5, a1 = 6; –14 1 2 1 64 1
128 1
256
an–1 2 5 4 1
n – 1 1 7
11-1

10. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
16. an = , a1 = 144;
17. an = an – 1 • , a1 = ;
18. an = n + 3; 15
19. an = ;
20. an = 3n + 1; 37
21. an = 4n – 1; 47
22. an = ; 3
23. an = n2 + 1; 145
24. recursive; 3, 9, 21, 45, 93
25. explicit; 0, 1, 3, 6, 10
26. explicit; –24, –21, –16, –9, 0
an – 1 4
27. recursive; –2, 6, –18, 54, –162
28. explicit; –6, –18, –38, –66, –102
29. explicit; 3, 9, 19, 33, 51
30. explicit; 5, 10, 15, 20, 25
31. recursive; 340, 323, 306, 289, 272
32. 15, 26, 40
33. 20, 23; an = 3n + 2; explicit or
an = an – 1 + 3, a1 = 5; recursive
34. 96, 192; an = 3 • 2n–1; explicit or
an = 2an – 1, a1 = 3; recursive
, 343; an = n3; explicit
, 16,384; an = 4n; explicit or
an = 4an – 1; a1 = 4, recursive 9 16 1 2 1 2 1 64 1
n – 1 1 13
n – 6 2
11-1

11. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
, 169; an = (n + 6)2; explicit or
an+1 = an + 2n + 13; a1 = 49, recursive
38. –1, 1; an = –1(an–1), a1 = –1; recursive or
an = ( –1)n; explicit
39. –1,– ; an = , a1 = –16; recursive
or an = ; explicit
40. –47, –40; an = an–1 + 7, a1 = –75;
recursive or an = –82 + 7n; explicit
41. –11, –19; an = an–1 – 8, a1 = 21;
recursive or an = 29 – 8n; explicit
42. an–2, an+2
43. Answers may vary. Sample: A recursive
formula requires that the previous term
be known to find a given term. An explicit
formula only requires the number of the term.
44. a–c. Answers may vary. Sample:
a. 1, –2, 4, –8, ...
b. an = –2(an–1), a1 = 1; an = (–2)n–1
c. –524,288
45. 26,677; 458,330; 210,066,388,901
46. 24, 78, 240, 726
47. 25, 36, 49, 64
48. 54, 128, 250, 432
, , ,
, , ,
51. a. 25 boxes
b. 110 boxes
c. 9 levels
an–1 2 1 2
–32 2n 16 5 25 6 36 7 49 8 5 6 6 7 7 8 8 9
11-1

12. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
52. an = 10 • 2n–1
53. an = –n – 4
54. an = –2 •
55. an = 1 + 4(n –1)
56. a. an = an–1 + 5, a1 = 25; an = n
b. $40
c. an = (an–1 + $20) • 1.005, a1 = $40.20
d. 6.5%
57. a. 15, 21
b. an = an–1 + n, a1 = 1
c. Yes; the formula yields the same
values as the recursive formula.
58. B
59. G
60. C
61. I
62. [2] an = an–1 + n2, a1 = 1;
Each term consists of the
previous term added to the
square of the number of
the term.
[1] incorrect formula OR
no explanation OR
incorrect explanation
n–1 1 2
11-1

13. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
63. (x + 4)2 + (y + 2)2 = 5
= 1
65. xy = 20
66. xy = 10
67. xy = 117
68. xy = 27
69. xy = 10
70. xy = 72
71. xy = –
72. xy = 100
(x + 1)2 36
(y + 2)2 36 1 4
11-1

14. Mathematical Patterns
ALGEBRA 2 LESSON 11-1
Describe each pattern. Find the next three terms.
2. 1, , , , . . . 2 3 4 9 8 27
1. 5, 15, 25, 35, . . .
Each new term is of the preceding term; 2 3 16 81 32
243 64
729 ,
Each new term is 10 more than the preceding term; 45, 55, 65
3. Write a recursive formula for the sequence 7, –1, –9, –17, Then find the next term.
an = an – 1 – 8, where a1 = 7; –25
4. Write an explicit formula for the sequence 1, , , , Then find a15 1 4 9 16
an = ; 1 n2
225
5. A recursive formula for a sequence is an = an – 1 + 2n, where a1 = 1. Write the first five terms of the sequence.
1, 5, 11, 19, 29
11-1

15. Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
(For help, go to Skills Handbook page 838.)
Describe the pattern in each sequence. Use at least one of the words add, subtract, or difference.
1. 10, 8, 6, 4, 2, 0, , 117, 134, 151, 168, . . .
3. , , , 2, – , – , – , –1, – , – , . . . 5 7 8 11 1 4 2 3
11-2

16. Arithmetic Sequences Solutions 1. 10, 8, 6, 4, 2, 0, . . .; subtract 2
ALGEBRA 2 LESSON 11-2
Solutions
1. 10, 8, 6, 4, 2, 0, . . .; subtract 2
2. 100, 117, 134, 151, 168, . . .; add 17
3. , , , 2, . . .; add
4. – , – , – , –1, – , – , . . .; subtract 5 7 8 11 1 4 2 3
11-2

17. Is the given sequence arithmetic?
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
Is the given sequence arithmetic?
a. 7, 10, 13, 16, . . .
7,           10,            13,            16
10 – 7 = 3 +3
13 – 10 = 3 +3
16 – 13 = 3 +3
The common difference is +3. This is an arithmetic sequence.
11-2

18. b. The sequence of dots in the “triangles” shown below.
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
(continued)
b. The sequence of dots in the “triangles” shown below.
3,            6,            10,            15
6 – 3 = 3 +3
10 – 6 = 4 +4
15 – 10 = 5 +5
There is no common difference. This is not an arithmetic sequence.
11-2

19. Find the 27th term of the sequence 75, 87, 99, . . . .
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
Suppose you have already saved $75 toward the purchase of a new CD player and speakers. You plan to save at least $12 a week from money you earn at a part-time job. In all, what is the minimum amount you will have after 26 weeks?
Find the 27th term of the sequence 75, 87, 99,
an = a1 + (n – 1)d Use the explicit formula.
a27 = 75 + (27 – 1)(12) Substitute a1 = 75, n = 27, and d = 12.
= 75 + (26)(12) Subtract within parentheses.
= Multiply.
= 387 Simplify.
After 26 weeks, you will have saved a minimum of $387.
11-2

20. Find the missing term of the arithmetic sequence 50, , 92.
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
Find the missing term of the arithmetic sequence 50, , 92.
arithmetic mean =
Write the average. 2
= Simplify the numerator.
142 2
= 71 Divide.
The missing term is 71.
11-2

21. Arithmetic Sequences pages 596–598 Exercises 1. no 2. yes; 10 3. no
ALGEBRA 2 LESSON 11-2
pages 596–598  Exercises
1. no
2. yes; 10
3. no
4. no
5. yes; 3
6. yes; –11
7. yes; 4
8. no
9. no
10. no
16. –159
17. –59
19. –146
21. –7.5
22. 21
23. 13
24. 16
25. –7
29. a11 or
31. 4
32.
33. 13
35. –19.5
a10 + a12 2 1 2
11-2

22. 43. The student assumed that the sequence was an = 2n–1.
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
37. –1
38.
39.
40.
41. 0
42. 2x + 1
43. The student assumed that
the sequence was an = 2n–1.
However, a1 = 20 = 1, not 0
as given in the problem.
44. a. Answers may vary. Sample:
25, 18, 11, 4, –3, –10,...; to find
the nth term, multiply n – 1
times (–7) and add to a1.
b. Answers may vary. Sample:
Start with the first term and continue
to subtract 7 for each term. For each
term, you subtract 7x (term number –1)
from the first term.
45. Answers may vary. Sample: An advantage
of a recursive formula is that only the
preceding term must be known to find
the next term; a disadvantage is that
many calculations may be required to find
a term. An advantage of an explicit formula
is that it is easy to find any term. 4 5
r + s 2
2r + s 2
11-2

23. 54. an = –5 + 1(n – 1); an = an–1 + 1, a1 = –5
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
46. 23
47. 15
49. 22
50. 6
51. 29
52. an = 2 + 2(n – 1); an = an–1 + 2, a1 = 2
53. an = 0 + 6(n – 1); an = an–1 + 6, a1 = 0
54. an = –5 + 1(n – 1); an = an–1 + 1, a1 = –5
55. an = –4 – 4(n – 1); an = an–1 – 4, a1 = –4
56. an = –2 + 7(n – 1); an = an–1 + 7, a1 = –2
57. an = 27 – 12(n – 1); an = an–1 – 12, a1 = 27
58. an = – (n – 1);
an = an– , a1 = –5
59. an = – (n – 1);
an = an–1 + 12, a1 = –32
60. an = 1 + (n – 1);
an = an–1 + , a1 = 1
61. an = (n – 1); an = an–1 + , a1 = 0
62. 6 min; 1 min
63. –4, –10, –16
, –0.8, –6.2
65. –8, –17, –-26
, 5,
67. 17, 17, 17 1 3 1 8 1 8 19 5 31 5
11-2

24. Arithmetic Sequences 73. 68. 681, 702, 723 69. –12.5, –8, –3.5
ALGEBRA 2 LESSON 11-2
, 702, 723
69. –12.5, –8, –3.5
70. a + 5, a + 9, a + 13
71. a. $20, $45, $75, $110, $150,
$195, $245, $300, $360,
$425, $495
b. an = an – 1 + $20 + $5(n – 1), a1 = $20
c. $495
72.
73.
min or 2 min 28 s
75. 54
76. 8
77. 21st term
78. 43rd term
79. a1 = –1, d = 3
80. a1 = –4, d = 4
81. a1 = 52, d = –10
11-2

25. 91. [2] The third term, 31, is the mean of the first
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
82. a1 = –21 , d = 4
83. a1 = –100.5, d = 22
84. a1 = –9, d = 2.2
85. 9k + 32
86. 21k – 43
87. B
88. I
89. B
90. H 1 4 1 4
91. [2] The third term, 31, is the mean of the first
and fifth terms. The second term, 23, is the
mean of the first and third terms. The fourth
term, 39, is the mean of the third and fifth terms.
[1] incomplete explanation OR
incorrect term values
92. [4] 696; a2 – a1 = 10 – 3 = 7, so the common
difference is 7. Use a1 = 3, d = 7, and
n = 100 in the formula an = a1 + (n – 1)d
to find a100.
[3] explanation correct, but computational error
[2] incomplete explanation
[1] correct number, but no explanation
11-2

26. Arithmetic Sequences 93. recursive; –2, –7, –12, –17, –22
ALGEBRA 2 LESSON 11-2
93. recursive; –2, –7, –12, –17, –22
94. explicit; 6, 18, 36, 60, 90
95. explicit; 0, 3, 8, 15, 24
96. recursive; –121, –108,–95, –82, –69
97. (0, – 5 ) and (0, )
98. (–4 2, 0) and (4 2, 0)
99. (1 – , 0) and ( , 0)
100. ( , 3) and (1 – , 3)
r =
6 2V 2 3
11-2

27. 4. Find the 25th term of the arithmetic sequence 26, 13, 0, –13, . . .
Arithmetic Sequences
ALGEBRA 2 LESSON 11-2
Is the given sequence arithmetic? If so, identify the common difference.
1. –9, 0, 9, 18, . . .
yes; 9
, 0.01, 0.001, , . . . no
, , , , . . . 1 6 3 2
yes; 1 6
4. Find the 25th term of the arithmetic sequence 26, 13, 0, –13, . . .
–286
Find the missing term of each arithmetic sequence.
, , 12 , . . . 1 2
5. 8, , 20, . . . 14 8
11-2

28. Find the next term in each sequence.
Geometric Sequences
ALGEBRA 2 LESSON 11-3
(For help, go to Lesson 11-3.)
Find the next term in each sequence.
1. 1, 2, 4, 8, , 168, 84, 42, . . .
3. 0.1, 1, 10, 100, , 300, 100, . . .
11-3

29. Geometric Sequences Solutions 1. 1, 2, 4, 8, 16 (multiply by 2)
ALGEBRA 2 LESSON 11-3
Solutions
1. 1, 2, 4, 8, 16 (multiply by 2)
2. 336, 168, 84, 42, 21 (divide by 2)
3. 0.1, 1, 10, 100, 1000 (multiply by 10)
4. 900, 300, 100, (divide by 3)
100 3
11-3

30.  –6  –6  –6 Geometric Sequences
ALGEBRA 2 LESSON 11-3
Is the given sequence geometric? If so, identify the common ratio.
a. 1, –6, 36, –216, . . .
1,           –6,           36,           –216
–6 ÷ 1 = –6
 –6
36 ÷ –6 = –6
 –6
216 ÷ 36 = –6
 –6
There is a common ratio of –6. This is a geometric sequence.
11-3

31.  2   Geometric Sequences b. 2, 4, 6, 8, . . . 2, 4, 6, 8 4 ÷ 2 = 2
ALGEBRA 2 LESSON 11-3
(continued)
b. 2, 4, 6, 8, . . .
2,            4,             6,             8
4 ÷ 2 = 2
 2  3 2
6 ÷ 4 =  4 3
8 ÷ 6 =
There is no common ratio. This is not a geometric sequence.
11-3

32. an = a1 • r n – 1 Use the explicit formula.
Geometric Sequences
ALGEBRA 2 LESSON 11-3
Suppose you have equipment that can enlarge a photo to 120% of its original size. A photo has a length of 10 cm. Find the length of the photo after 5 enlargements at 120%.
You need to find the 6th term of the geometric sequence 10, 12, 14.4, . . .
an = a1 • r n – 1 Use the explicit formula.
a6 = 10 • – 1 Substitute a1 = 10, n = 6, and r = 1.20.
= 10 • Simplify the exponent.
Use a calculator.
After five enlargements of 120%, the photo has a length of about 25 cm.
11-3

33. geometric mean = 150,000 • 188, 160 Use the definition.
Geometric Sequences
ALGEBRA 2 LESSON 11-3
A family purchased a home for $150,000. Two years later the home was valued at $188,160. If the value of the home is increasing geometrically, how much was the home worth after one year?
geometric mean = ,000 • 188, 160 Use the definition.
= 28,224,000,000 Multiply.
= 168,000 Take the square root.
11-3

34. Geometric Sequences 12. no
ALGEBRA 2 LESSON 11-3
12. no
13. an = 5 • (–3)n – 1; 5, –15, 45, –135, 405
14. an = • 10n – 1; , 0.237,
2.37, 23.7, 237
15. an = ; , , , ,
16. an = 0.5n – 1; 1, 0.5, 0.25, 0.125,
17. an = 100(–20)n – 1; 100; –2000; 40,000;
–800,000; 16,000,000
18. an = 7 • 1n – 1; 7, 7, 7, 7, 7
19. an = 1024(0.5)n – 1; 1024, 512, 256, 128, 64
20. an = 4(0.1)n – 1; 4, 0.4, 0.04, 0.004,
21. an = 10(–1)n – 1; 10, –10, 10, –10, 10
pages 603–605  Exercises
1. yes; 2; 16, 32
2. no
3. yes; –2; 16, –32
4. yes; –1; –1, 1
5. yes; 0.4; 0.256,
6. yes; 0.1; ,
7. yes; – ; , –
8. no
9. yes; 1.5; ,
10. yes; –5; 1250, –6250
11. yes; 6; –1296, –7776 1 2 2 3
n – 1 1 2 1 3 2 9 4 27 8 81 1 3 2 9 2 27
11-3

35. 41. a–d. Answers may vary. Sample: a. 3 and 12; 6 b. 3, 6, 12; 2
Geometric Sequences
ALGEBRA 2 LESSON 11-3
35. geometric; 2, 2
36. neither; 25, 36
, 2187, 729
, 22.5, 67.5
39. 10, 8, 6.4
40. –6.64, –11.02, –18.30
41. a–d. Answers may vary. Sample:
a. 3 and 12; 6
b. 3, 6, 12; 2
c. 768
d. 48; 5th term
43. 12,288
,432
24.
27. 6
28. geometric; 720, 1440
29. arithmetic; 125, 150
30. geometric; 3, –3
31. arithmetic; 50, 55
32. geometric; –80, 160
33. geometric; 0.125,
34. neither; 20, 26 4 15
11-3

36. 54. Both the common difference and the common
Geometric Sequences
ALGEBRA 2 LESSON 11-3
,326,592
46. 12,884,901,888
47. 3(4n – 1)
48. 4
49. 16
51. 10
52. –
53. –
54. Both the common difference and the common
ratio are used to find the next term in a sequence,
but a common difference is added and a common
ratio is multiplied.
55. $142.79, $613.59, $28.62, $58.92, $105.82, $262.94
56. a. d, d, d, d, d, d
b. Yes; the common ratio is .
c. an = an – 1 • , a1 = d
57. Both arithmetic and geometric sequence explicit
formulas use the first term a1, n – 1, and a common
term. The recursive formulas both use an – 1 and
a common term. 1 2 1 4 1 8 1 16 1 32 1 2 1 2 1 6 2 3
11-3

37. e. The common ratio is less than one, so the graph is decreasing.
Geometric Sequences
ALGEBRA 2 LESSON 11-3
58. a. 5000, 3750, , , b.
c cm3 d.
e. The common ratio is less than one,
so the graph is decreasing.
59. 7
61. A
62. F
63. A
64. C
65. B
66. [2] geometric mean = • 16
= = 8; arithmetic mean
= = = 10; The
arithmetic mean is greater.
[1] correct answer with no
work shown 3 4
4 + 16 2 20 2 3 4
11-3

38. [3] appropriate methods, but with one computational error
Geometric Sequences
ALGEBRA 2 LESSON 11-3
70. an = –2 + (–11)(n – 1);
an = an – 1 – 11, a1 = –2
71. x2 + y2 = 9
72. (x + 3)2 + (y – 1)2 = 25
67. [4] Since a4 = a3 • r = a2 • r 2 =
a1• r 3, 192 = 3 • r 3. So r 3 = 64,
and r = = 4. Then a2 = a1• r
= 3 • 4 = 12 and a3 =
a2 • r = 12 • 4 = 48.
[3] appropriate methods, but with
one computational error
[2] correct answer, but with
minimal explanation
[1] correct answer, but with
no explanation
68. an = –3 + 3(n – 1);
an = an – 1 + 3, a1 = –3
69. an = 17 + (–9)(n – 1);
an = an – 1 – 9, a1 = 17 3
11-3

39. Geometric Sequences 73. (x – 1)2 + (y – 1)2 = 4 74. x = –3 75. x = –1
ALGEBRA 2 LESSON 11-3
73. (x – 1)2 + (y – 1)2 = 4
74. x = –3
75. x = –1
76. x = 0, 1
11-3

40. 5. Find the missing term for the geometric sequence 3, , 48 . . . 12
Geometric Sequences
ALGEBRA 2 LESSON 11-3
Is the given sequence geometric? If so, identify the common ratio and find the next two terms.
1. 1, 2, 6, 12, . . .
2. 2, 1, 0.5, 0.25, . . . no
yes; 0.5; 0.125,
3. –9, 81, –729, 6561, . . .
yes; –9, –59,049, 531,441
4. Write the explicit formula for the geometric sequence for which a1 = 7 and r = . Then generate the first five terms. 1 3
an = 7 • ; 7, , , , 1 3
n – 1 7 9 27 81
5. Find the missing term for the geometric sequence 3, , 12
11-3

41. Write an explicit formula for each sequence.
Arithmetic Series
ALGEBRA 2 LESSON 11-4
(For help, go to Lesson 11-1.)
Find each sum.
2. –17 + (–13) + (–9) + (–5) + (–1) + 3
Write an explicit formula for each sequence.
3. 4, 6, 8, 10, 12, , 4, 7, 10, 13, 16, . . .
5. –17, –23, –29, –35, , 1, –8, –17, . . .
11-4

42. Arithmetic Series Solutions
ALGEBRA 2 LESSON 11-4
Solutions
= (2 + 8) + ( ) + 5 = = 25
2. –17 + (–13) + (–9) + (–5) + (–1) + 3 = [–17 + (–13)] + [–9 + (–1)] + [–5 + 3] = –30 + (–10) + (–2) = –42
3. 4, 6, 8, 10, 12, . . .
an = a1 + (n – 1)d
= 4 + (n – 1)2
= 4 + 2n – 2
= 2 + 2n
4. 1, 4, 7, 10, 13, 16, . . .
an = a1 + (n – 1)d
= 1 + (n – 1)3
= 1 + 3n – 3
= –2 + 3n
5. –17, –23, –29, –35, . . .
an = a1 + (n – 1)d
= –17 + (n – 1)(–6)
= –17 – 6n + 6
= –11 – 6n
6. 10, 1, –8, –17, . . .
an = a1 + (n – 1)d
= 10 + (n – 1)(–9)
= 10 – 9n + 9
= 19 – 9n
11-4

43. The sum of the terms of the sequence is 119.
Arithmetic Series
ALGEBRA 2 LESSON 11-4
Use the finite sequence 5, 9, 13, 17, 21, 25, 29. Write the related series. Evaluate the series.
= 119
Related series
Add to evaluate.
The sum of the terms of the sequence is 119.
11-4

44. Define: Let Sn = total number of blocks,
Arithmetic Series
ALGEBRA 2 LESSON 11-4
A staircase uses same-size cement blocks arranged 4 across, as shown below. Find the total number of blocks in the staircase.
Relate:
sum of
the series is
number of terms 2
times
the first
term
plus
the last
Define: Let Sn = total number of blocks,
and let n = the number of stairs.
Then a1 = the number of blocks in the first stair,
and an = the number of blocks in the last stair.
11-4

45. Write: Sn = ( a1 + an ) Use the formula.
Arithmetic Series
ALGEBRA 2 LESSON 11-4
(continued) n 2
Write: Sn = ( a1 + an ) Use the formula.
= (4 + 20) Substitute n = 5, a1 = 4, an = 20. 5 2
= 2.5(24) Simplify.
= 60 Multiply.
There are 60 blocks in the stairs.
11-4

46.  Arithmetic Series Use the summation notation to write the series
ALGEBRA 2 LESSON 11-4
Use the summation notation to write the series
for 50 terms.
8 • 1 = 8, 8 • 2 = 16, 8 • 3 = 24, . . .   The explicit formula for the sequence is 8n.
= n The lower limit is 1 and the upper limit is 50.  50
n = 1
11-4

47.  Arithmetic Series Use the series (–2n + 3).
ALGEBRA 2 LESSON 11-4  4
n = 1
Use the series (–2n + 3).
a. Find the number of terms in the series.
Since the values of n are 1, 2, 3, and 4, there are four terms in the series.
b. Find the first and last terms in the series.
The first term of the series –2n + 3 = –2(1) + 3 = 1.
The last term of the series –2n + 3 = –2(4) + 3 = –5.
11-4

48.  Arithmetic Series c. Evaluate the series.
ALGEBRA 2 LESSON 11-4
(continued)
c. Evaluate the series.  4
n = 1
(–2n + 3) = (–2(1) + 3) + (–2(2) + 3) + (–2(3) + 3) +
(–2(4) + 3) Substitute.
= 1 + (–1) + (–3) + (–5)   Simplify within parentheses.
= –8 Add.
The sum of the series is –8.
11-4

49. 2. (–5) + (–15) + (–25) + (–35) + (–45); –125
Arithmetic Series
ALGEBRA 2 LESSON 11-4
pages 610–612  Exercises
; 84
2. (–5) + (–15) + (–25) + (–35) + (–45); –125
; 585
(–0.25) + (–0.5) + (–0.75); –0.75
; 109.5
; 54.6
7. 32
8. –48
9. 264
12. –146 n
14. (n + 7)
15. (n + 4)
16. (3n – 2) n
18. –3n 4
n = 1 8
n = 1 7
n = 1 11
n = 1 15
n = 1 5
n = 1
11-4

50. 390 chairs middle, 930 chairs total
Arithmetic Series
ALGEBRA 2 LESSON 11-4
19. 5, 1, 9; 25
20. 5, –3, –11; –35
21. 6, 4, –1; 9
22. 5, 0, 0.8; 2
23. 9, , ;72
24. 6, 15, 10; 75
25. sequence; infinite
26. sequence; finite
27. series; finite
28. series; infinite
29. sequence; infinite
30. series; finite
31. a. 270 chairs on each side,
390 chairs middle, 930 chairs total
b. each side: (n + 3); middle: (n + 9)
c. $46,950
32. a. 8; the formula for the corresponding
sequence is 3n + 7. Solving 3n + 7 = 31
for n shows that n = 8.
b. 164
33. a. 91
b. 83 20 20
n = 1
n = 1 8 3 40 3
11-4

51. would not be an integer for 110 cans or 140 cans. 35. 110 36. –765
Arithmetic Series
ALGEBRA 2 LESSON 11-4
34. a. an = n + 1
b (n + 1)
c. 18 cans
d. No; no; 13 rows have 104 cans,
14 rows have 119 cans, 15 rows
have 135 cans, and 16 rows have
152 cans. The number of rows
would not be an integer for 110
cans or 140 cans.
36. –765
38. –22 9
n = 1
39. –0.6
40. 1,000,500
41. a. No; (6) = 225,
which is less than 500.
b. Answers may vary. Sample:
Pro: spreading the cost over
several years, con: calculators
purchased first may be outdated
by the time 500 calculators have
been purchased; check
students’ work.
42. a–d. Check students’ work.
44. –200
45. 34
11-4

52. ellipse: x- and y-axes; domain: –6 x 6; range: –2 3 y 2 3 n – 1
Arithmetic Series
ALGEBRA 2 LESSON 11-4
46. 12,884,901,888
47. 10x + 45y
48. 45x – 210y
49. B
50. G
51. C
52. A
53. C
54. B
55. an = 1(2)n – 1; 1, 2, 4
56. an = 1(5)n – 1; 1, 5, 25
57. an = –1(–1)n – 1; –1, 1, –1 3 2
58. an = ; 3, ,
59. an = –7(0.1)n – 1; –7, –0.7, –0.07
60. an = 20(–0.5)n – 1; 20, –10, 5
61.
ellipse: x- and y-axes; domain: –6 x 6;
range: – y
n – 1 9 2 27 4
< –
< –
< –
< –
11-4

53. circle: all lines through the origin; domain: –2 x 2; range: –2 y 2
Arithmetic Series
ALGEBRA 2 LESSON 11-4
64.
65.
66.
x + 3
x – 4
c – 2
c – 5
z z + 20
z – 1
63.
circle: all lines through the origin;
domain: –2 x 2; range: –2 y 2
< –
62.
hyperbola: x- and y-axes; domain: x –5 or x 5;
range: all real numbers
>
11-4

54.     Arithmetic Series
ALGEBRA 2 LESSON 11-4  8
n = 1
1. Write an addition expression of the finite series represented by
10n.
Write each finite arithmetic series using summation notation. Then find the sum of each series.  6
n = 1
( n); 642  6
n = 1
(15 – 5n); –15
(–5) + (–10) + (–15)
4. Write the arithmetic series to the given term using summation notation. Then find the sum of the series.  6
n = 1
3n; 15, 150
; 100th
11-4

55. Find each sum or difference. 1. 100 + 50 + 25 + + 2. 3 + 9 + 27 + 81
Geometric Series
ALGEBRA 2 LESSON 11-5
(For help, go to Lesson 9-5.)
Find each sum or difference.
3. –2 + 4 – – –5 – 10 – 20 – 40
Simplify each fraction. 25 2 4
1 – 1 5 3 – 16
2 +
11-5

56. Geometric Series Solutions
ALGEBRA 2 LESSON 11-5
= = = = 193
= (3 + 27) + (9 + 81) = = 120
3. –2 + 4 – – 32 = (4 + 16) – ( ) = 20 – 42 = –22
4. –5 – 10 – 20 – 40 = –(5 + 20) – ( ) = –25 – 50 = –75
Solutions 25 2 25 4 50 4 25 4 75 4 3 4 3 4
11-5

57. Solutions (continued)
Geometric Series
ALGEBRA 2 LESSON 11-5
Solutions (continued)
1 – 1 5 3 4 5 1 3
5. = = ÷ = • = or 2
6. = = 1 ÷ = 1 • = or 1
7. = = = = ÷ = • = =
8. = = ÷ = • = or 6 4 5 1 3 4 5 3 1 12 5 2 5 1 4
1 – 1 3 4 3 4 4 3 1 3 1 2 3 – 4
1 • 3
2 • 3
1 • 2
3 • 2 – 1 4 3 6 2 – 1 4 1 6 4 1 6 1 4 1 6 4 1 4 6 2 3 1 16
2 + 3 33 16 1 3 33 16 1 3 33 16 3 1 99 16 3 16
11-5

58. Use the formula to evaluate the series 5 + 15 + 45 + 135 + 405 + 1215.
Geometric Series
ALGEBRA 2 LESSON 11-5
Use the formula to evaluate the series
The first term is 5, and there are six terms in the series.
The common ratio is = = = = = 3 15 5 45
135
405
1215
So a1 = 5, r = 3, and n = 6.
Sn = Write the formula.
a1 (1 – r n)
1 – r
= Substitute a1 = 5, r = 3, and n = 6.
5 (1 – 36)
1 – 3
= = 1820 Simplify.
–3640 –2
The sum of the series is 1820.
11-5

59. Write the formula for the sum of a geometric series.
ALGEBRA 2 LESSON 11-5
The Floyd family starts saving for a vacation that is one year away. They start with $125. Each month they save 8% more than the previous month. How much money will they have saved 12 months later?
Relate:
Sn =
a1 (1 – r n)
1 – r
Write the formula for the sum of
a geometric series.
Define: Sn = total amount saved
a1 = 125 Initial amount.
r = 1.08 Common ratio.
n = 12 Number of months.
Write:
S12 =
125 ( 1 – )
1 – 1.08
Use a calculator.
Substitute.
The amount of money the Floyd’s will have saved will be $
11-5

60. Geometric Series
ALGEBRA 2 LESSON 11-5
Decide whether each infinite geometric series diverges or converges. Then determine whether the series has a sum.  
n = 1 2 3 n a. b
a1 = = , a2 = = 2 3 1 4 9
a1 = 2, a2 = 6
r = ÷ = 4 9 2 3
r = 6 ÷ 2 = 3
Since | r | < 1, the series converges, and the series has a sum.
Since | r | 1, the series diverges, and the series does not have a sum.
>
11-5

61. The total distance that the pendulum swings through is 200 in.
Geometric Series
ALGEBRA 2 LESSON 11-5
The weight at the end of a pendulum swings through an arc of 30 inches on its first swing. After that, each successive swing is 85% of the length of the previous swing. What is the total distance the weight will swing by the time it comes to rest? The largest arc the pendulum swings through is on the first swing of 30 in., so a1 = 30.
S = Use the formula. a1
1 – r
= Substitute. 30
1 – 0.85
= 200 Simplify.
The total distance that the pendulum swings through is 200 in.
11-5

62. Geometric Series pages 616–619 Exercises 1. 255 2. 1456 3. 381 4. 3647
ALGEBRA 2 LESSON 11-5
pages 616–619  Exercises
1. 255
3. 381
5. –10,235
6. – 7.
8. –1640
9. converges; has a sum
10. converges; has a sum
11. converges; has a sum
12. diverges; no sum
13. diverges; no sum
14. converges; has a sum
15. diverges; no sum
16. converges; has a sum
17. diverges; no sum
19. 1
20.
21.
22. 9
23.
24. geometric; 2046
25. arithmetic; 420
26. geometric; –1,627,605
27. geometric;
28. arithmetic; 500,500
29. geometric; 121.5 9 5
1111 6
255
256 5 6 9 2
11-5

63. Geometric Series 32. 30. a. 33. 4 34. 35. no sum 36. 0.83 37. no sum
ALGEBRA 2 LESSON 11-5
30. a.
4, 16, 64 b
c. 5460
31. a. 20, 18, 16.2, 14.58 b
c. S = = 200
d. Check students’ work.
32.
33. 4
34.
35. no sum
37. no sum
38. a. 2
b  . .
c  . .
d. 225 cm
39. Check students’ work. 5 4 3 4 20
1 – 0.9
11-5

64. 50. a. all integers greater than or equal to 1
Geometric Series
ALGEBRA 2 LESSON 11-5
46. diverges
47. converges;
48. a. 70th swing
b. 10,000 cm
49. a. S = = = b.
50. a. all integers greater than
or equal to 1
b. 10; 18; 24.4; 29.52; 33.62; 36.89;
39.51; 41.61; 43.29; 44.63;
c. 50
40. a. No; the sum of a series of positive
numbers will be positive.
b. Your classmate did not check if |r |
were less than 1.
41.
42. 10
43. (b); (a) yields $26,000; using the formula
for finding the sum of a finite geometric
series, (b) yields $1,342,
44. a. Answers may vary. Sample:
The student used r – 1 instead of 1 – r
in the formula for the sum of an infinite
geometric series. b.
45. converges; 1 7 8
1 – 1 7 3 7 1 2
11-5

65. Geometric Series 51. 64. 61. (–1, 0), x = 1; 52. 65. 53. 66. 54. 5
ALGEBRA 2 LESSON 11-5
51.
52.
53.
54. 5
58. –825
60. (0, 4), y = –4; 4 7
64.
65.
66.
61. (–1, 0), x = 1;
62. 0, – , y = ;
63.
10(2y + 3)
(x + 3)(x – 3) 2 3
x2 + 6x + 4
(x + 6)(x – 6) 2 3
–3(5d + 16)
(d + 3)(d – 3) 9 4 9 4
7c – 4
2c2
11-5

66. 1. Find the sum of the geometric series. 3 – 30 + 300 – 3000 + 30,000
ALGEBRA 2 LESSON 11-5
1. Find the sum of the geometric series.
3 – – ,000
27,273
2. Find the sum of the first twelve terms of the geometric series. Round your answer to the nearest thousandth.
31.992
3. How can you tell whether an infinite geometric series converges or diverges?
Check the common ratio r. If | r | < 1, the series converges. Otherwise it diverges.
4. Find the sum of the infinite geometric series. 32
5. Find the sum of the infinite geometric series. 5 3 9
22.5
11-5

67. Find the area of the rectangle with the given length and width.
Area Under a Curve
ALGEBRA 2 LESSON 11-6
(For help, go to Skills Handbook page 847.)
Find the area of the rectangle with the given length and width.
1. = 4 ft, w = 1 ft = 5.5 m, w = 0.5 m
3. = 6.2 cm, w = 0.1 cm = 9 in., w = 3 in. 1 2 5 8
11-6

68. Area Under a Curve Solutions 1. A = w = (4 ft)(1 ft) = 4 ft2
ALGEBRA 2 LESSON 11-6
Solutions
1. A = w = (4 ft)(1 ft) = 4 ft2
2. A = w = (5.5 m)(0.5 m) = 2.75 m2
3. A = w = (6.2 cm)(0.1 cm) = 0.62 cm2
4. A = w = (9 in.)(3 in.) = = = in.2 1 2 5 8 19 2 29 8
551 16 7 16
11-6

69. a. What does the area under the curve represent?
Area Under a Curve
ALGEBRA 2 LESSON 11-6
The curve shown below approximates the speed of a car during a 12-minute drive.
a. What does the area under the curve represent?
Area = • minutes   Use dimensional analysis.
feet
minutes
= feet Simplify.
The area under the curve approximates the total distance traveled by the car.
11-6

70. Value of curve at upper edge of each rectangle.
Area Under a Curve
ALGEBRA 2 LESSON 11-6
(continued)
b. Use the inscribed rectangle 2 units wide to estimate the area under the curve.
Total area
2(2000) + 2(3000) + 2(3500) + 2(4000) + 2(3000) + 2(2000) = 35,000
Value of curve at upper edge of each rectangle.
Width of each rectangle
The indicated area is 35,000 units2.
11-6

71.  Area Under a Curve Estimate the area under the curve
ALGEBRA 2 LESSON 11-6
Estimate the area under the curve
ƒ(x) = –0.5x2 + 6 for the domain
0 x 2 by evaluating the sum A.
< –
< –
A = (0.5)ƒ(an), where a1 = 0.5, a2 = 1, a3 = 1.5, a4 = 2.  4
n = 1
A = 0.5ƒ(0.5) + 0.5ƒ(1) + 0.5ƒ(1.5) + 0.5ƒ(2)   Add the areas of the rectangles.
= 0.5( ) total area = width of each rectangle • sum of the heights
= 0.5(20.25) Add within parentheses.
= Simplify.
The indicated area is about units2.
11-6

72. Step 1: Input the equation. Adjust the window values.
Area Under a Curve
ALGEBRA 2 LESSON 11-6
Step 1: Input the equation. Adjust the window values.
Use a graphing calculator to graph
ƒ(x) = –x2 + 4x + 5. Find the area under
the curve for the domain 1 x 4.
< –
< –
Step 2: Access the ƒ(x)dx feature from the CALC menu.
Step 3: Use the lower limit of x = 1.
Step 4: Use the upper limit of x = 4.
The area under the curve between x = 1 and x = 4 is 24 units2.
11-6

73. Area Under a Curve pages 625–627 Exercises 1. total produced
ALGEBRA 2 LESSON 11-6
pages 625–627  Exercises
1. total produced
2. amount of growth
3. miles
4. distance traveled
5. total price
units2
units2
units2
9. A = ƒ(an )
a. 0.5 units2
b. 2.5 units2
10. A = ƒ(an )
a. 5 units2
b. 9 units2
11. A = g (an )
a. 3 units2
b. 7 units2
12. A = ƒ(an )
a. 3 unit2 2
13. A = ƒ(an )
a units2
b units2
14. A = h (an )
a. 5 units2
b. 25 units2
15. A = ƒ(an )
a unit2
b units2 2
n = 1
n = 1 2 3 2 1 3 2
n = 1
n = 1 2 2
n = 1 2
n = 1
n = 1
11-6

74. Area Under a Curve 16. A = 1h (an ) a. 5 units2 b. 9 units2
ALGEBRA 2 LESSON 11-6
16. A = h (an )
a. 5 units2
b. 9 units2
17. A = ƒ(an )
units2
units2
units2
units2
units2
units2 2
24.
8.75 units2
25.
43 units2
26.
28.75 units2
27.
50 units2
28.
units2
29.
19.5 units2
n = 1 2
n = 1
11-6

75. approximates the area because it is between the other measures
Area Under a Curve
ALGEBRA 2 LESSON 11-6
30. a.
b. 3 units2 c.
12 units2
d. 7.5 units2; the mean best
approximates the area because
it is between the other measures
known to be larger and smaller
than the actual value.
31. a. Answers may vary. Sample:
0.37 miles
b. Answers may vary. Sample: Using
inscribed rectangles will result in
an estimate smaller than the
actual number.
32. 9 units2
units2
units2
units2
11-6

76. c. The estimate in (a) is closer to the actual area because there
Area Under a Curve
ALGEBRA 2 LESSON 11-6
units2
units2
units2
units2
units2
41. Check students’ work.
42. a units2
b. 14 units2
c. The estimate in (a) is closer
to the actual area because there
is less area between the curve
and the rectangles when more
rectangles are used.
43. a.
b. The area under both curves is
the same, 3 units2, over the interval
– x ; this is true because
the amount of area above y = x3 + 1
and below y = 1 on the left side of the
y-axis is equal to the area below
y = x3 + 1and above y = 1 on the
right side of the y-axis.
44. 6 units2
< –
< – 1 4 1 4
11-6

77. 50. [2] The mean might be a better estimate
Area Under a Curve
ALGEBRA 2 LESSON 11-6
9x2 25
45. a. y = – b.
23.56 units2
c units2
d. Check students’ work.
46. A
47. G
48. B
49. G
50. [2] The mean might be a better estimate
since it is between the estimate using
inscribed rectangles, which is too low,
and the estimate using circumscribed
rectangles, which is too high.
[1] incomplete explanation
51. exists
52. exists
53. – = 1; x2 16
y 2 9
11-6

78. Area Under a Curve 54. – = 1 55. – = 1 56. –3 57. 58. 11-6 x2 y 2 25 1
ALGEBRA 2 LESSON 11-6 x2 25
y 2 1
54. – = 1
55. – = 1
56. –3
57.
58. x2 10
y 2 16 3 14
7 ± 9
11-6

79. Area Under a Curve
ALGEBRA 2 LESSON 11-6
1. Use circumscribed rectangles to approximate the area under the graph of the function ƒ(x) = 5 – x2, for x Use rectangles 1 unit wide. 1 4
< –
< –
about units2
2. Use inscribed rectangles to approximate the area under the graph of the function ƒ(x) = x, for –3 x Use rectangles 0.5 units wide.
< –
< –
about 2.96 units2
3. Evaluate a sum to approximate the area under the graph of the function ƒ(x) = 3x2, for x Use inscribed rectangles 2 units wide.
< –
< –
about 720 units2
4. Use a graphing calculator to find the area under the graph of y = x3 – 2x2 + 3, x
< –
< –
11.25 units2
11-6

80. Sequences and Series Page 632 13. 2, –4, 8, –16, 32
ALGEBRA 2 CHAPTER 11
Page 632
13. 2, –4, 8, –16, 32
14. 3, 10, 17, 24, 31
15. –100, –20, –4, – ,–
16. 19, 15, 11, 7, 3
17. 1
18. 4
19.
20.
21. 36
22. arithmetic; 156
23. geometric; 6250
24. arithmetic; –4.9
1. an = an – 1 + 6, a1 = 7, an = 1 + 6n; 73
2. an = an – 1 • 2, a1 = 10, an = 10 • 2n–1; 20,480
3. a. savings = (months – 1)
b. $75
4. arithmetic; 59
5. arithmetic; 51
6. geometric; 98,415
7. 8
8. 6
9. Check students’ work.
10. geometric; r =
11. arithmetic; d = –3
12. geometric; r = 2 4 5 4 25 5 9 2 3 1 3
11-A

81. 29. If the absolute value of the common ratio is less than 1,
Sequences and Series
ALGEBRA 2 CHAPTER 11
25. 5, 4, 16; 50
26. 8, , ; 24
27. 7, 2.8, 7.6; 36.4
28. 5, –2, –32; –22
29. If the absolute value of the
common ratio is less than 1,
then it will converge;
check students’ work.
30. $
cm, 36 cm, 32.4 cm,
29.16 cm
32. miles
33. pounds
34. dollars
37.
9 units2
38. Answers may vary. Sample:
You could find the sum of the
circumscribed rectangles and
the sum of the inscribed
rectangles and average
the numbers.
35.
2 units2
36.
4 units2 2 3 16 3
11-A