10-2 Arithmetic Sequences and Series
Write an equation for the nth term of each
arithmetic sequence.
32. CCSS STRUCTURE José averaged 123 total pins per game in his bowing league this season. He is
taking bowling lessons and hopes to bring his average
up by 8 pins each new season.
29. a. Write an equation to represent the nth term of the
sequence.
SOLUTION: b. If the pattern continues, during what season will
José average 187 per game?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
SOLUTION: a. Given d = 8 and a 1 = 123.
Find the nth term.
Use the value of a 1 to find the nth term.
b. Substitute 187 for a n and solve for n.
Therefore José’s average will be 187 pins per game
th
in the 9 season.
32. CCSS STRUCTURE José averaged 123 total pins per game in his bowing league this season. He is
taking bowling lessons and hopes to bring his average
up by 8 pins each new season.
c. Sample answer: No; there are a maximum of 300
points in a bowling game, so it would be impossible
for the average to continue to climb indefinitely.
a. Write an equation to represent the nth term of the
sequence.
Find the arithmetic means in each sequence.
b. If the pattern continues, during what season will
José average 187 per game?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
SOLUTION: a. Given d = 8 and a 1 = 123.
Find the nth term.
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b. Substitute 187 for a n and solve for n.
38. SOLUTION: Here a 1 = 182 and a 7 = 104.
Therefore, the missing numbers are (182 – 13) orPage 1
169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13)
or 130, and (130 – 13) or 117.
c. Sample answer: No; there are a maximum of 300
points in a bowling game, so it would be impossible
the averageSequences
to continueand
to climb
indefinitely.
10-2for
Arithmetic
Series
Find the arithmetic means in each sequence.
42. the first 300 even natural numbers
SOLUTION: Here a 1 = 2 and a 300 = 300.
38. SOLUTION: Here a 1 = 182 and a 7 = 104.
n = 300
Find the sum.
Therefore, the missing numbers are (182 – 13) or
169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13)
or 130, and (130 – 13) or 117.
46. n = 19, a n = 154, d = 8
Find the sum of each arithmetic series.
SOLUTION: Find the value of a 1.
41. the first 100 odd natural numbers
SOLUTION: Here a 1 = 1 and a 100 = 199.
n = 100
Find the sum.
Find the sum.
42. the first 300 even natural numbers
SOLUTION: Here a 1 = 2 and a 300 = 300.
Find the first three terms of each arithmetic
series.
54. a 1 = 19, n = 44, S n = 9350
SOLUTION: Find the value of a n.
n = 300
Find the sum.
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10-2 Arithmetic Sequences and Series
Find the first three terms of each arithmetic
series.
54. a 1 = 19, n = 44, S n = 9350
SOLUTION: Find the value of a n.
Therefore, the first three terms are 19, 28 and 37.
55. a 1 = –33, n = 36, S n = 6372
SOLUTION: Find the value of a n.
Find the value of d.
Find the value of d.
Therefore, the first three terms are –33, –21 and –9.
Therefore, the first three terms are 19, 28 and 37.
55. a 1 = –33, n = 36, S n = 6372
SOLUTION: Find the value of a n.
56. PRIZES A radio station is offering a total of $8500
in prizes over ten hours. Each hour, the prize will
increase by $100. Find the amounts of the first and
last prize.
SOLUTION: Given n = 10, d = 100 and S 10 = 8500.
Find the value of a 1.
Find the value of d.
Find the value of a 10.
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theSequences
first three terms
are –33, –21 and –9.
10-2Therefore,
Arithmetic
and Series
56. PRIZES A radio station is offering a total of $8500
in prizes over ten hours. Each hour, the prize will
increase by $100. Find the amounts of the first and
last prize.
Find the sum of each arithmetic series.
57. SOLUTION: Given n = 10, d = 100 and S 10 = 8500.
SOLUTION: There are 16 – 1 + 1 or 16 terms, so n = 16.
Find the value of a 1.
Find the sum.
Find the value of a 10.
Therefore,
.
Find the sum of each arithmetic series.
58. 57. SOLUTION: There are 13 – 4 + 1 or 10 terms, so n = 10.
SOLUTION: There are 16 – 1 + 1 or 16 terms, so n = 16.
Find the sum.
Find the sum.
Therefore,
.
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Manual - Powered by Cognero .
Therefore,
Page 4
59. Therefore,
.
Therefore,
10-2 Arithmetic Sequences and Series
58. .
60. SOLUTION: There are 13 – 4 + 1 or 10 terms, so n = 10.
SOLUTION: There are 12 – 0 + 1 or 13 terms, so n = 13.
Find the sum.
Find the sum.
Therefore,
.
Therefore,
.
Use the given information to write an equation
that represents the nth term in each arithmetic
sequence.
59. SOLUTION: There are 16 – 5 + 1 or 12 terms, so n = 12.
63. The 100th term of the sequence is 245. The common
difference is 13.
SOLUTION: Given a 100 = 245, d = 13 and n = 100.
Find the sum.
Find the value of a 1.
Substitute the values of a 1 and d to find the nth term.
Therefore,
.
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SOLUTION: Page 5
64. The eleventh term of the sequence is 78. The
common difference is –9.
Therefore,
.
10-2 Arithmetic Sequences and Series
Use the given information to write an equation
that represents the nth term in each arithmetic
sequence.
63. The 100th term of the sequence is 245. The common
difference is 13.
65. The sixth term of the sequence is –34. The 23rd term
is 119.
SOLUTION: Given a 6 = –34 and a 23 = 119.
SOLUTION: Given a 100 = 245, d = 13 and n = 100.
Therefore, there are (23 – 6 + 1) or 18 terms
between –34 and 119.
Find the common difference of the series with a 1 = –
Find the value of a 1.
34 and a 18 = 119.
Substitute the values of a 1 and d to find the nth term.
Find the value of a 1.
Substitute the values of a 1 and d to find the nth term.
64. The eleventh term of the sequence is 78. The
common difference is –9.
SOLUTION: Given a 11 = 78, d = –9 and n = 11.
66. The 25th term of the sequence is 121. The 80th term
is 506.
Find the value of a 1.
SOLUTION: Given a 25 = 121 and a 80 = 506.
Substitute the values of a 1 and d to find the nth term.
Therefore, there are (80 – 25 + 1) or 56 terms
between 121 and 506.
Find the common difference of the series with a 1 =
121 and a 56 = 506.
sixth term
of the
65. The Manual
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is –34. The 23rd term
Page 6
is 119.
10-2 Arithmetic Sequences and Series
66. The 25th term of the sequence is 121. The 80th term
is 506.
SOLUTION: Given a 25 = 121 and a 80 = 506.
67. CCSS MODELING The rectangular tables in a
reception hall are often placed end-to-end to form
one long table. The diagrams below show the number
of people who can sit at each of the table
arrangements.
Therefore, there are (80 – 25 + 1) or 56 terms
between 121 and 506.
Find the common difference of the series with a 1 =
a. Make drawings to find the next three numbers as
tables are added one at a time to the arrangement.
121 and a 56 = 506.
b. Write an equation representing the nth number in
this pattern.
Find the value of a 1.
c. Is it possible to have seating for exactly 100 people
with such an arrangement? Explain.
SOLUTION: a. For each increase in the number of table, the
number of people who can sit is increased by 4. That
is, the common difference is 4.
Therefore, the next three numbers are (10 + 4) or 14,
(14 + 4) or 18 and (18 + 4) or 22.
Substitute the values of a 1 and d to find the nth term.
67. CCSS MODELING The rectangular tables in a
reception hall are often placed end-to-end to form
one long table. The diagrams below show the number
of people who can sit at each of the table
arrangements.
a. Make drawings to find the next three numbers as
tables are added one at a time to the arrangement.
b. Write an equation representing the nth number in
this pattern.
eSolutions
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c. IsManual
it possible
to have
seating
for exactly 100 people
with such an arrangement? Explain.
b. Substitute a 1 = 6 and d = 4 in
.
c. No; there is no whole number n for which
.
70. SPORTS While training for cross country, Silvia
plans to run 3 miles per day for the first week, and
then increase the distance by a half mile each ofPage
the 7
following weeks.
day.
c. No; there is no whole number n for which
10-2 Arithmetic. Sequences and Series
c. Sample answer: No; eventually the number of
miles per day will become unrealistic.
70. SPORTS While training for cross country, Silvia
plans to run 3 miles per day for the first week, and
then increase the distance by a half mile each of the
following weeks.
Find the value of x.
72. a. Write an equation to represent the nth term of the
sequence.
SOLUTION: There are x – 3 + 1 or x – 2 terms, so n = x – 2.
b. If the pattern continues, during which week will
she be running 10 miles per day?
c. Is it reasonable for this pattern to continue
indefinitely? Explain.
Find the sum.
SOLUTION: a. Given a 1 = 3 and d = 0.5.
Find the nth term.
Equate the sum with the given value and solve for x.
b. Substitute 10 for a n in
and solve for n.
During 15th week, she will be running 10 miles per
day.
c. Sample answer: No; eventually the number of
miles per day will become unrealistic.
The value of x should be positive. Therefore, x = 18.
Find the value of x.
73. 72. SOLUTION: There are x – 3 + 1 or x – 2 terms, so n = x – 2.
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Find the sum.
SOLUTION: There are x – 5 + 1 or x – 4 terms, so n = x – 4.
Find the sum.
Page 8
The value of x should be positive. Therefore, x = 18.
10-2 Arithmetic Sequences and Series
73. SOLUTION: There are x – 5 + 1 or x – 4 terms, so n = x – 4.
The value of x should be positive. Therefore, x = 16.
74. CCSS CRITIQUE Eric and Juana are determining
the formula for the nth term for the sequence –11, –
2, 7, 16, … . Is either of them correct? Explain your
reasoning.
Find the sum.
Equate the sum with the given value and solve for x.
SOLUTION: Sample answer: Eric; Juana missed the step of
multiplying d by n – 1.
75. REASONING If a is the third term in an arithmetic
sequence, b is the fifth term, and c is the eleventh
term, express c in terms of a and b.
SOLUTION: Given a 3 = a, a 5 = b and a 11 = c.
Find the common difference.
The value of x should be positive. Therefore, x = 16.
74. CCSS CRITIQUE Eric and Juana are determining
the formula for the nth term for the sequence –11, –
2, 7, 16, … . Is either of them correct? Explain your
reasoning.
Find the value of a 1.
Find the value of c in terms of a and b.
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Page 9
SOLUTION: Sample answer: Eric; Juana missed the step of
d by
n – 1. and Series
10-2multiplying
Arithmetic
Sequences
75. REASONING If a is the third term in an arithmetic
sequence, b is the fifth term, and c is the eleventh
term, express c in terms of a and b.
SOLUTION: Given a 3 = a, a 5 = b and a 11 = c.
Find the common difference.
Find the value of a 1.
Find the value of c in terms of a and b.
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