10-2 Arithmetic Sequences and Series
Find the indicated term of each arithmetic sequence.
1. a 1 = 14, d = 9, n = 11
SOLUTION: Write an equation for the nth term of each arithmetic sequence.
3. 13, 19, 25, …
SOLUTION: 19 – 13 = 6
25 – 19 = 6
The common difference d is 6.
a 1 = 13
Find the arithmetic means in each sequence.
5. SOLUTION: Here a 1 = 6 and a 5 = 42.
Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33.
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Find the sum of each arithmetic series.
7. the first 50 natural numbers
Page 1
10-2 Arithmetic Sequences and Series
Find the arithmetic means in each sequence.
5. SOLUTION: Here a 1 = 6 and a 5 = 42.
Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33.
Find the sum of each arithmetic series.
7. the first 50 natural numbers
SOLUTION: 9. a 1 = 12, a n = 188, d = 4
SOLUTION: Find the sum of the series.
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Find the first three terms of each arithmetic series.
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10-2 Arithmetic Sequences and Series
9. a 1 = 12, a n = 188, d = 4
SOLUTION: Find the sum of the series.
Find the first three terms of each arithmetic series.
11. a 1 = 8, a n = 100, S n = 1296
SOLUTION: Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16.
13. MULTIPLE CHOICE Find
A 45
B 78
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C 342
D 410
Page 3
Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16.
10-2 Arithmetic Sequences and Series
13. MULTIPLE CHOICE Find
A 45
B 78
C 342
D 410
SOLUTION: There are 12 – 1 + 1 or 12 terms, so n = 12.
Find the sum.
Option C is the correct answer.
Find the indicated term of each arithmetic sequence.
15. a 1 = –12, n = 66, d = 4
SOLUTION: 17. a 15 for –5, –12, –19, …
SOLUTION: –12 – (–5) = –7
–19 – (–12) = –7
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The common difference d is –7.
a = –5
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10-2 Arithmetic Sequences and Series
17. a 15 for –5, –12, –19, …
SOLUTION: –12 – (–5) = –7
–19 – (–12) = –7
The common difference d is –7.
a 1 = –5
19. a 24 for 8.25, 8.5, 8.75, …
SOLUTION: 8.5 – 8.25 = 0.25
8.75 – 8.5 = 0.25
The common difference d is 0.25.
a 1 = 8.25
Write an equation for the nth term of each arithmetic sequence.
21. 31, 17, 3, …
SOLUTION: 17 – 31 = –14
3 – 17 = –14
The common difference d is –14.
a 1 = 31
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10-2 Arithmetic Sequences and Series
Write an equation for the nth term of each arithmetic sequence.
21. 31, 17, 3, …
SOLUTION: 17 – 31 = –14
3 – 17 = –14
The common difference d is –14.
a 1 = 31
23. a 7 = 21, d = 5
SOLUTION: Use the value of a 1 to find the nth term.
25. a 5 = 1.5, d = 4.5
SOLUTION: Use the value of a 1 to find the nth term.
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10-2 Arithmetic Sequences and Series
25. a 5 = 1.5, d = 4.5
SOLUTION: Use the value of a 1 to find the nth term.
27. a 6 = 22, d = 9
SOLUTION: Use the value of a 1 to find the nth term.
29. SOLUTION: Manual - Powered by Cognero
eSolutions
Use the value of a 1 to find the nth term.
Page 7
10-2 Arithmetic Sequences and Series
29. SOLUTION: Use the value of a 1 to find the nth term.
31. SOLUTION: Use the value of a 1 to find the nth term.
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Find the arithmetic means in each sequence.
33. Page 8
10-2 Arithmetic Sequences and Series
31. SOLUTION: Use the value of a 1 to find the nth term.
Find the arithmetic means in each sequence.
33. SOLUTION: Here a 1 = 24 and a 6 = –1.
Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4.
35. SOLUTION: Here a 1 = –28 and a 6 = 7.
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Therefore, the missing numbers are (24 – 5) or 19, (19 – 5) or 14, (14 – 5) or 9, and (9 – 5) or 4.
10-2 Arithmetic Sequences and Series
35. SOLUTION: Here a 1 = –28 and a 6 = 7.
Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0.
37. SOLUTION: Here a 1 = –12 and a 7 = –66.
Therefore, the missing numbers are (–12 – 9) or –21, (–21 – 9) or –30, (–30 – 9) or –39, (–39 – 9) or –48, and (–48
– 9) or –57.
Find the sum of each arithmetic series.
39. the first 100 even natural numbers
SOLUTION: Here a 1 = 2 and a 100 = 200.
Find the sum.
eSolutions
Manual
by Cognero
41. the first
100- Powered
odd natural
numbers
SOLUTION: Page 10
10-2 Arithmetic Sequences and Series
41. the first 100 odd natural numbers
SOLUTION: Here a 1 = 1 and a 100 = 199.
n = 100
Find the sum.
43. –18 + (–15) + (–12) + … + 66
SOLUTION: –15 – (–18) = 3
–12 – (–15) = 3
The common difference is 3.
Find the value of n.
Find the sum.
45. a 1 = –16, d = 6, n = 24
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SOLUTION: Page 11
10-2 Arithmetic Sequences and Series
45. a 1 = –16, d = 6, n = 24
SOLUTION: 47. CONTESTS The prizes in a weekly radio contest began at $150 and increased by $50 for each week that the
contest lasted. If the contest lasted for eleven weeks, how much was awarded in total?
SOLUTION: Given, a 1 = 150, d = 50 and n = 11.
Find the value of a 11.
Find the sum.
A cash prizes totaled $4400 for the eleven week contest.
Find the first three terms of each arithmetic series.
49. a 1 = 48, a n = 180, S n = 1368
SOLUTION: Find the value of n.
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Find the value of d.
Page 12
cash prizes Sequences
totaled $4400
forSeries
the eleven week contest.
10-2AArithmetic
and
Find the first three terms of each arithmetic series.
49. a 1 = 48, a n = 180, S n = 1368
SOLUTION: Find the value of n.
Find the value of d.
So, the sequence is 48, 60, 72, …
51. n = 28, a n = 228, S n = 2982
SOLUTION: Find the value of a 1.
Find the value of d.
Therefore, the first three terms are –15, –6 and 3.
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53. n = 30, a n = 362, S n = 4770
Page 13
So, the sequence is 48, 60, 72, …
10-2 Arithmetic Sequences and Series
51. n = 28, a n = 228, S n = 2982
SOLUTION: Find the value of a 1.
Find the value of d.
Therefore, the first three terms are –15, –6 and 3.
53. n = 30, a n = 362, S n = 4770
SOLUTION: Find the value of a 1.
Find the value of d.
Therefore, the first three terms are –44, –30 and –16.
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55. a = –33, n = 36, S = 6372
Page 14
theSequences
first three terms
are –15, –6 and 3.
10-2Therefore,
Arithmetic
and Series
53. n = 30, a n = 362, S n = 4770
SOLUTION: Find the value of a 1.
Find the value of d.
Therefore, the first three terms are –44, –30 and –16.
55. a 1 = –33, n = 36, S n = 6372
SOLUTION: Find the value of a n.
Find the value of d.
Therefore, the first three terms are –33, –21 and –9.
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Find the sum of each arithmetic series.
Page 15
theSequences
first three terms
are –44, –30 and –16.
10-2Therefore,
Arithmetic
and Series
55. a 1 = –33, n = 36, S n = 6372
SOLUTION: Find the value of a n.
Find the value of d.
Therefore, the first three terms are –33, –21 and –9.
Find the sum of each arithmetic series.
57. SOLUTION: There are 16 – 1 + 1 or 16 terms, so n = 16.
Find the sum.
Therefore,
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.
Page 16
theSequences
first three terms
are –33, –21 and –9.
10-2Therefore,
Arithmetic
and Series
Find the sum of each arithmetic series.
57. SOLUTION: There are 16 – 1 + 1 or 16 terms, so n = 16.
Find the sum.
Therefore,
.
59. SOLUTION: There are 16 – 5 + 1 or 12 terms, so n = 12.
Find the sum.
Therefore,
.
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61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end of
the first month and $25 more each additional month for 12 months. How much does she pay in total after the 12
Therefore,
.
10-2 Arithmetic Sequences and Series
59. SOLUTION: There are 16 – 5 + 1 or 12 terms, so n = 12.
Find the sum.
Therefore,
.
61. FINANCIAL LITERACY Daniela borrowed some money from her parents. She agreed to pay $50 at the end of
the first month and $25 more each additional month for 12 months. How much does she pay in total after the 12
months?
SOLUTION: Given a 1 = 50, d = 25 and n = 12.
Find the sum.
She pays $2250.
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.
SOLUTION: Given a 100 = 245, d = 13 and n = 100.
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Find the value of a 1.
Page 18
pays $2250.
10-2She
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.
SOLUTION: Given a 100 = 245, d = 13 and n = 100.
Find the value of a 1.
Substitute the values of a 1 and d to find the nth term.
65. The sixth term of the sequence is –34. The 23rd term is 119.
SOLUTION: Given a 6 = –34 and a 23 = 119.
Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.
Find the common difference of the series with a 1 = –34 and a 18 = 119.
Find the value of a 1.
Substitute the values of a 1 and d to find the nth term.
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10-2 Arithmetic Sequences and Series
65. The sixth term of the sequence is –34. The 23rd term is 119.
SOLUTION: Given a 6 = –34 and a 23 = 119.
Therefore, there are (23 – 6 + 1) or 18 terms between –34 and 119.
Find the common difference of the series with a 1 = –34 and a 18 = 119.
Find the value of a 1.
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.
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.
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10-2 Arithmetic Sequences and Series
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.
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.
b. Substitute a 1 = 6 and d = 4 in
.
c. No; there is no whole number n for which
.
69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how
many years will he have a salary of $100,000 per year?
SOLUTION: Given a 1 = 28000, d = 4000 and a n = 100000.
Substitute the values of a 1, a n and d and solve for n.
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No; there isSequences
no whole number
n for which
10-2c.Arithmetic
and Series
.
69. SALARY Terry currently earns $28,000 per year. If Terry expects a $4000 increase in salary every year, after how
many years will he have a salary of $100,000 per year?
SOLUTION: Given a 1 = 28000, d = 4000 and a n = 100000.
Substitute the values of a 1, a n and d and solve for n.
So he will have a salary of $100,000 per year after the 19th year.
71. MUTIPLE REPRESENTATIONS Consider
a. TABULAR Make a table of the partial sums of the series for 1 ≤ k ≤ 10.
b. GRAPHICAL Graph (k, partial sum).
c. GRAPHICAL Graph f (x) = x2 + 3x on the same grid.
d. VERBAL What do you notice about the two graphs?
e. ANALYTICAL What conclusions can you make about the relationship between quadratic functions and the sum
of arithmetic series?
2
f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x + 8x.
SOLUTION: a.
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10-2 Arithmetic Sequences and Series
b.
c.
d. Sample answer: The graphs cover the same range. The domain of the series is the natural numbers, while the
domain of the quadratic function is all real numbers, 0 ≤ x ≤ 10.
e . Sample answer: For every partial sum of an arithmetic series, there is a corresponding quadratic function that
shares the same range.
f.
Find the value of x.
73. SOLUTION: There are x – 5 + 1 or x – 4 terms, so n = x – 4.
Find the sum.
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shares the same range.
10-2f.Arithmetic Sequences and Series
Find the value of x.
73. SOLUTION: There are x – 5 + 1 or x – 4 terms, so n = x – 4.
Find the sum.
Equate the sum with the given value and solve for x.
The value of x should be positive. Therefore, x = 16.
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.
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The value of x should be positive. Therefore, x = 16.
10-2 Arithmetic Sequences and Series
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.
77. CHALLENGE Find S n for (x + y) + (x + 2y) + (x + 3y) + … .
SOLUTION: (x + 2y) – (x + y) = y
(x + 3y) – (x + 2y) = y
The common difference is y.
Find the sum.
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10-2 Arithmetic Sequences and Series
77. CHALLENGE Find S n for (x + y) + (x + 2y) + (x + 3y) + … .
SOLUTION: (x + 2y) – (x + y) = y
(x + 3y) – (x + 2y) = y
The common difference is y.
Find the sum.
79. WRITING IN MATH Compare and contrast arithmetic sequences and series.
SOLUTION: Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common
difference. An arithmetic series is the sum of the terms of an arithmetic sequence.
81. PROOF Derive a sum formula that does not include a 1.
SOLUTION: General sum formula
a n = a 1 + (n – 1)d Formula for nth term
a n – (n – 1)d = a 1 Subtract (n – 1)d from both sides.
Substitution
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Simplify.
Page 26
SOLUTION: Sample answer: An arithmetic sequence is a list of terms such that any pair of successive terms has a common
AnSequences
arithmetic series
is the sum of the terms of an arithmetic sequence.
10-2difference.
Arithmetic
and Series
81. PROOF Derive a sum formula that does not include a 1.
SOLUTION: General sum formula
a n = a 1 + (n – 1)d Formula for nth term
a n – (n – 1)d = a 1 Subtract (n – 1)d from both sides.
Substitution
Simplify.
83. SAT/ACT The measures of the angles of a triangle form an arithmetic sequence. If the measure of the smallest
angle is 36°, what is the measure of the largest angle?
A 54°
B 75°
C 84°
D 90°
E 97°
SOLUTION: The sum of the interior angle of a triangle is 180°.
Since the measures of the angles of a triangle form an arithmetic sequence and the smallest angle is 36°, the other two angles are 36° + d and 36° + 2d.
36 + 36 + d + 36 + 2d = 180
d = 24
Therefore, the largest angle is 36° + 48° = 84°.
Option C is the correct answer.
85. The expression
is equivalent to
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A
Page 27
36 + 36 + d + 36 + 2d = 180
d = 24
Therefore, the largest angle is 36° + 48° = 84°.
10-2 Arithmetic Sequences and Series
Option C is the correct answer.
85. The expression
is equivalent to
A
B
C
D
SOLUTION: Option A is the correct answer.
Determine whether each sequence is arithmetic. Write yes or no.
87. –6, 4, 14, 24, …
SOLUTION: Since there is a common difference between the consecutive terms, this is an arithmetic sequence.
89. 10, 8, 5, 1, ...
SOLUTION: Since there is no common difference between the consecutive terms, this is not an arithmetic sequence.
Solve each system of inequalities by graphing.
91. SOLUTION: eSolutions Manual - Powered by Cognero
Page 28
SOLUTION: there is Sequences
no commonand
difference
10-2Since
Arithmetic
Seriesbetween the consecutive terms, this is not an arithmetic sequence.
Solve each system of inequalities by graphing.
91. SOLUTION: 93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =
km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k 1
and k 2 are attached in a series, the resulting spring constant k is found by the equation
a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of
8 centimeters per gram, find the resultant spring constant.
b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in
this context?
SOLUTION: a. Given k 1 = 12 and k 2 = 8.
Substitute the values and evaluate the value of k.
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10-2 Arithmetic Sequences and Series
93. PHYSICS The distance a spring stretches is related to the mass attached to the spring. This is represented by d =
km, where d is the distance, m is the mass, and k is the spring constant. When two springs with spring constants k 1
and k 2 are attached in a series, the resulting spring constant k is found by the equation
a. If one spring with constant of 12 centimeters per gram is attached in a series with another spring with constant of
8 centimeters per gram, find the resultant spring constant.
b. If a 5-gram object is hung from the series of springs, how far will the springs stretch? Is this answer reasonable in
this context?
SOLUTION: a. Given k 1 = 12 and k 2 = 8.
Substitute the values and evaluate the value of k.
b. Substitute 4.8 and 5 for k and m respectively in the equation d = km.
The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40
cm. The object would have to stretch the combined springs less than it would stretch either of the springs
individually.
Graph each function. State the domain and range.
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95. Page 30
The answer is reasonable. The object would stretch the first spring 60 cm and would stretch the second spring 40
cm. The object would have to stretch the combined springs less than it would stretch either of the springs
10-2individually.
Arithmetic Sequences and Series
Graph each function. State the domain and range.
95. SOLUTION: Graph the function.
The function is defined for all values of x.
Therefore, the domain is D = {all real numbers}.
The value of the f (x) tends to 3 as x tends to –∞.
The value of the f (x) tends to ∞ as x tends to ∞.
Therefore, the range of the function is R ={f (x) | f (x) > 3}.
Solve each equation. Round to the nearest ten-thousandth.
x
97. 5 = 52
SOLUTION: n+2
99. 3
= 14.5
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10-2 Arithmetic Sequences and Series
n+2
99. 3
= 14.5
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Page 32