10-2 Arithmetic Sequences and Series

Transcription

Determine the common difference, and find the next four terms of each arithmetic sequence.
1. 20, 17, 14, …
SOLUTION: First, find the common difference.
17 – 20 = –3
14 – 17 = –3
The common difference is –3. Add –3 to the third term to find the fourth term, and so on.
14 + (–3) = 11
11 + (–3) = 8
8 + (–3) = 5
5 + (–3) = 2
Therefore, the next four terms are 11, 8, 5, and 2.
2. 3, 16, 29, …
16 – 3 = 13
29 – 16 = 13
The common difference is 13. Add 13 to the third term to find the fourth term, and so on.
29 + 13 = 42
42 + 13 = 55
55 + 13 = 68
68 + 13 = 81
3. 117, 108, 99, …
108 –117 = –9
99 – 108 = –9
99 – 9 = 90
90 – 9 = 81
81 – 9 = 72
72 – 9 = 63
4. −83, −61, −39, …
–83 – 61 = 22
–61 – 39 = 22
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–39 + 22 = –17
Page 1
90 – 9 = 81
81 – 9 = 72
72 – 9 = 63
4. −83, −61, −39, …
–83 – 61 = 22
–61 – 39 = 22
–39 + 22 = –17
–17 + 22 = 5
5 + 22 = 27
27 + 22 = 49
Therefore, the next four terms are –17, 5, 27, and 49.
5. −3, 1, 5, …
1 – (–3) = 4
5– 1=4
5+4=9
9 + 4 = 13
13 + 4 = 17
17 + 4 = 21
Therefore, the next four terms are 9, 13, 17, and 21
6. 4, 21, 38, …
21 – 4 = 17
38 – 21 = 17
38 + 17 = 55
55 + 17 = 72
72 + 17 = 89
89 + 17 =106
7. −4.5, −9.5, −14.5, …
–9.5 –(–4.5) = –5
–14.5 – (–9.5) = –5
–14.5 – 5 = –19.5
–19.5 – 5 = –24.5
–24.5 – 5 = –29.5
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55 + 17 = 72
72 + 17 = 89
89 + 17 =106
7. −4.5, −9.5, −14.5, …
–9.5 –(–4.5) = –5
–14.5 – (–9.5) = –5
–14.5 – 5 = –19.5
–19.5 – 5 = –24.5
–24.5 – 5 = –29.5
–29.5 – 5 = –34.5
Therefore, the next four terms are –19.5, –24.5, –29.5, and –34.5.
8. −97, −29, 39, …
–29 – (–97) = 68
39 – (–29) = 68
39 + 68 = 107
107 + 68 = 175
175 + 68 = 243
243 + 68 = 311
9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band
member, the second row has 3 band members, the third row has 5 band members, and so on.
a. Find the number of band members in the 8th row.
b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.
SOLUTION: a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,
so a 1 = 1, a 2 = 3, and a 3 = 5. Find the common difference.
3– 1=2
5– 3=2
The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term.
5+2=7
7+2=9
9 + 2 = 11
11 + 2 = 13
13 + 2 = 15
So, there are 15 band members in the 8th row.
b. For an explicit formula, substitute a 1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.
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107 + 68 = 175
175 + 68 = 243
243 + 68 = 311
9. MARCHING BAND A marching band begins its performance in a pyramid formation. The first row has 1 band
member, the second row has 3 band members, the third row has 5 band members, and so on.
a. Find the number of band members in the 8th row.
b. Write an explicit formula and a recursive formula for finding the number of band members in the nth row.
SOLUTION: a. The first row has 1 band member, the second row has 3 band members, and the third row has 5 band members,
so a 1 = 1, a 2 = 3, and a 3 = 5. Find the common difference.
3– 1=2
5– 3=2
The common difference is 2. Add 2 to the third term to find the fourth term, and so on to find the eighth term.
5+2=7
7+2=9
9 + 2 = 11
11 + 2 = 13
13 + 2 = 15
So, there are 15 band members in the 8th row.
b. For an explicit formula, substitute a 1 = 1 and d = 2 in the formula for the nth term of an arithmetic sequence.
For the recursive formula, state the first term a 1 and then indicate that the next term is the sum of the previous
term a n – 1 and d.
a 1 = 1, a n = a n – 1 + 2
Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.
10. 2, 5, 8, …
5– 2=3
8– 5=3
For an explicit formula, substitute a 1 = 2 and d = 3 in the formula for the nth term of an arithmetic sequence.
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10-2 aArithmetic
and Series
= 1, a = a Sequences
+2
1
n
n –1
Find both an explicit formula and a recursive formula for the nth term of each arithmetic sequence.
10. 2, 5, 8, …
5– 2=3
8– 5=3
a 1 = 2, a n = a n – 1 + 3
11. −6, 5, 16, …
5 – (–6) = 11
16 – 5 = 11
For an explicit formula, substitute a 1 = –6 and d = 11 in the formula for the nth term of an arithmetic sequence.
a 1 = –6, a n = a n – 1 + 11
12. −9, −16, −23, …
–16 – (–9) = –7
–23 – (–16) = –7
For an explicit formula, substitute a 1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.
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10-2 aArithmetic
and Series
= –6, a = a Sequences
+ 11
1
n
n –1
12. −9, −16, −23, …
–16 – (–9) = –7
–23 – (–16) = –7
For an explicit formula, substitute a 1 = –9 and d = –7 in the formula for the nth term of an arithmetic sequence.
a 1 = –9, a n = a n – 1 – 7
13. 4, 19, 34, …
19 – 4 =15
34 – 19 = 15
a 1 = 4, a n = a n – 1 + 15
14. 25, 11, −3, …
11 – 25 = –14
–3 – 11 = –14
For an explicit formula, substitute a 1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.
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10-2 aArithmetic
and Series
+ 15
1
n
n –1
14. 25, 11, −3, …
11 – 25 = –14
–3 – 11 = –14
For an explicit formula, substitute a 1 = 25 and d = –14 in the formula for the nth term of an arithmetic sequence.
a 1 = 25, a n = a n – 1 – 14
15. 7, −3.5, −14, …
–3.5 – 7 = –10.5
–14 – (–3.5) = –10.5
For an explicit formula, substitute a 1 = 7 and d = –10.5 in the formula for the nth term of an arithmetic sequence.
a 1 = 7, a n = a n – 1 – 10.5
16. −18, 4, 26, …
4 – (–18 ) = 22
26 – 4 = 22
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10-2 aArithmetic
and Series
– 10.5
1
n
n –1
16. −18, 4, 26, …
4 – (–18 ) = 22
26 – 4 = 22
a 1 = –18, a n = a n – 1 + 22
17. 1, 37, 73, …
37 – 1 = 36
73 – 37 = 36
a 1 = 1, a n = a n – 1 + 36
Find the specified value for the arithmetic sequence with the given characteristics.
18. If a 14 = 85 and d = 9, find a 1.
SOLUTION: Substitute a n = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.
19. Find d for −24, −31, −38, …
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10-2 aArithmetic
and Series
+ 36
1
n
n –1
Find the specified value for the arithmetic sequence with the given characteristics.
18. If a 14 = 85 and d = 9, find a 1.
SOLUTION: Substitute a n = 85, n = 14, and d = 9 into the formula for the nth term of an arithmetic sequence.
19. Find d for −24, −31, −38, …
SOLUTION: Find the difference between two pairs of consecutive terms.
–31 – (–24) = –7
–38 – (–31) = –7
Therefore, d = –7.
20. If a n = 14, a 1 = −36, and d = 5, find n.
SOLUTION: Substitute a n = 14, a 1 = –36, and d = 5 into the formula for the nth term of an arithmetic sequence.
21. If a 1 = 47 and d = −5, find a 12.
SOLUTION: Substitute a 1 = 47, n = 12, and d = –5 into the formula for the nth term of an arithmetic sequence.
22. If a 22 = 95 and a 1 = 11, find d.
SOLUTION: Substitute a 1 = 11, n = 22, and a 22 = 95 into the formula for the nth term of an arithmetic sequence.
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1
22. If a 22 = 95 and a 1 = 11, find d.
SOLUTION: Substitute a 1 = 11, n = 22, and a 22 = 95 into the formula for the nth term of an arithmetic sequence.
23. Find a 6 for 84, 5, −74, …
SOLUTION: 5 – 84 = –79
–74 – 5 = –79
Substitute a 1 = 84, n = 6, and d = –79 into the formula for the nth term of an arithmetic sequence.
24. If a n = −20, a 1 = 46, and d = −11, find n.
SOLUTION: Substitute a 1 = 46, d = –11, and a n = −20 into the formula for the nth term of an arithmetic sequence.
25. If a 35 = −63 and a 1 = 39, find d.
SOLUTION: Substitute a 1 = 39, n = 35, and a 35 = −63 into the formula for the nth term of an arithmetic sequence.
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26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let a n represent the number of
25. If a 35 = −63 and a 1 = 39, find d.
SOLUTION: Substitute a 1 = 39, n = 35, and a 35 = −63 into the formula for the nth term of an arithmetic sequence.
26. CONSTRUCTION Each 8-foot section of a wooden fence contains 14 pickets. Let a n represent the number of
pickets in n sections.
a. Find the first five terms of the sequence.
b. Write a recursive formula for the sequence in part a.
c. If 448 pickets were used to fence in the customer’s backyard, how many feet of fencing was used?
SOLUTION: a. If each section contains 14 pickets, a 1 = 14 and d = 14.
14 + 14 = 28
28 + 14 = 42
42 + 14 = 56
56 + 14 = 70
Therefore, the first 5 terms of the sequence are 14, 28, 42, 56, 70.
b. Substitute a 1 = 14 and d = 14 into the formula for the nth term of an arithmetic sequence.
c. Find the number of sections n. Substitute a n = 448,
a 1 = 14, and d = 14 into the formula you found in part b.
Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.
Find the indicated arithmetic means for each set of nonconsecutive terms.
27. 3 means; 19 and −5
SOLUTION: The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a 5.
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First, find the common difference using a 5 = –5, a 1 = 19, and n = 5.
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Because each section is 8 feet wide and there are 32 sections, 8(32) or 256 feet of fencing was used.
Find the indicated arithmetic means for each set of nonconsecutive terms.
27. 3 means; 19 and −5
SOLUTION: The sequence will resemble 19, __?__ , __?__ , __?__ , –5. Note that –5 is the fifth term of the sequence or a 5.
First, find the common difference using a 5 = –5, a 1 = 19, and n = 5.
Next, determine the arithmetic means by using d = –6.
19 + (–6) = 13
13 + (–6) = 7
7 + (–6) = 1
Therefore, a sequence with three arithmetic means between 19 and –5 is 19, 13, 7, 1, –5.
28. 5 means; –62 and −8
SOLUTION: The sequence will resemble –62, __?__ , __?__ , __?__ , __?__ , __?__ , –8. Note that –8 is the seventh term of
the sequence or a 7.
First, find the common difference using a 7 = –8, a 1 = –62, and n = 7.
Next, determine the arithmetic means by using d = 9.
–62 + 9 = –53
–53 + 9 = –44
–44 + 9 = –35
–35 + 9 = –26
–26 + 9 = –17
Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.
29. 4 means; 3 and 88
SOLUTION: The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence
or a 6.
First, find the common difference using a 6 = 88, a 1 = 3, and n = 6.
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–44 + 9 = –35
–35 + 9 = –26
–26 + 9 = –17
Therefore, a sequence with five arithmetic means between –62 and −8 is –62, –53, –44, –35, –26, –17, –8.
29. 4 means; 3 and 88
SOLUTION: The sequence will resemble 3, __?__ , __?__ , __?__ , __?__ , 88. Note that 88 is the sixth term of the sequence
or a 6.
3 + 17 = 20
20 + 17 = 37
37 + 17 = 54
54 + 17 = 71
Therefore, a sequence with four arithmetic means between 3 and 88 is 3, 20, 37, 54, 71, 88.
30. 8 means; −5.5 and 23.75
SOLUTION: The sequence will resemble –5.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 23.75. Note
that 23.75 is the tenth term of the sequence or a 10.
First, find the common difference using a 10 = 23.75, a 1 = –5.5, and n = 10.
Next, determine the arithmetic means by using d = 3.25.
–5.5 + 3.25 = –2.25
–2.25 + 3.25 = 1
1 + 3.25 = 4.25
4.25 + 3.25 = 7.5
7.5 + 3.25 = 10.75
10.75 + 3.25 = 14
14 + 3.25 = 17.25
17.25 + 3.25 = 20.5
Therefore, a sequence with eight arithmetic means between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14,
17.25, 20.5, 23.75.
31. 7 means; −4.5 and 7.5
SOLUTION: Page
–4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5
is 13
the ninth term of the sequence or a 9.
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The
sequence
willbyresemble
14 + 3.25 = 17.25
17.25 + 3.25 = 20.5
a sequence
with eight
arithmetic
10-2 Therefore,
Arithmetic
Sequences
and
Seriesmeans between –5.5 and 23.75 is –5.5, –2.25, 1, 4.25, 7.5, 10.75, 14,
17.25, 20.5, 23.75.
31. 7 means; −4.5 and 7.5
SOLUTION: The sequence will resemble –4.5, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , 7.5. Note that 7.5 is
the ninth term of the sequence or a 9.
First, find the common difference using a 9 = 7.5, a 1 = –4.5, and n = 9.
Next, determine the arithmetic means by using d = 1.5.
–4.5 + 1.5 = –3
–3 + 1.5 = –1.5
–1.5 + 1.5 = 0
0 + 1.5 = 1.5
1.5 + 1.5 = 3
3 + 1.5 = 4.5
4.5 + 1.5 = 6
Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.
32. 10 means; 6 and 259
SOLUTION: The sequence will resemble 6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__
259. Note that 259 is the eleventh term of the sequence or a 12.
6 + 23 or 29
39 23 or 52
52 23 or 75
7 + 23 or 98
98 + 23 or 121
121 + 23 or 144
144 + 23 or 167
167 + 23 or 190
190 + 23 or 213
Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213,
236.
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Find a quadratic model for each sequence.
33. Page 14
1.5 + 1.5 = 3
3 + 1.5 = 4.5
4.5 + 1.5 = 6
Therefore, a sequence with seven arithmetic means between –4.5 and 7.5 is −3, −1.5, 0, 1.5, 3, 4.5, 6.
32. 10 means; 6 and 259
SOLUTION: The sequence will resemble 6, __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__ , __?__
259. Note that 259 is the eleventh term of the sequence or a 12.
6 + 23 or 29
39 23 or 52
52 23 or 75
7 + 23 or 98
98 + 23 or 121
121 + 23 or 144
144 + 23 or 167
167 + 23 or 190
190 + 23 or 213
Therefore, a sequence with ten arithmetic means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213,
236.
33. 12, 19, 28, 39, 52, 67, …
SOLUTION: 2
The nth term can be represented by a quadratic equation of the form a n = an + bn + c. Substitute values for a n
and n into the equation.
This yields a system of linear equations in three variables.
You can use a graphing calculator to solve for a, b, and c. First, write the augmented matrix that corresponds to
the system.
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Next,
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graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH
167 + 23 or 190
190 + 23 or 213
a sequence
with ten and
arithmetic
means between –6 and 259 is 29, 52, 75, 98, 121, 144, 167, 190, 213,
10-2 Therefore,
Arithmetic
Sequences
Series
236.
33. 12, 19, 28, 39, 52, 67, …
SOLUTION: 2
the system.
Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX] MATH
Therefore, a = 1, b = 4, and c = 7. Substituting these values in the equation for a n, the model for the sequence is a n
2
= n + 4n + 7c.
34. −11, −9, −5, 1, 9, 19, …
SOLUTION: 2
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the system.
to solve for a, b, and c. First, write the augmented matrix that corresponds Page
to 16
2
10-2 =Arithmetic
n + 4n + 7c. Sequences and Series
34. −11, −9, −5, 1, 9, 19, …
SOLUTION: 2
the system.
Next, enter the matrix into your graphing calculator, and use the rref( feature under the 2nd [MATRIX ]MATH
Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for a n, the model for the sequence
2
is a n = n – n – 11.
35. 8, 3, −6, −19, −36, −57, …
SOLUTION: 2
You
can -use
a graphing
calculator
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the system.
to solve for a, b, and c. First, write the augmented matrix that corresponds Page
to 17
Therefore, a = 1, b = –1, and c = –11. Substituting these values in the equation for a n, the model for the sequence
2
10-2 isArithmetic
and Series
a = n – n – Sequences
11.
n
35. 8, 3, −6, −19, −36, −57, …
SOLUTION: 2
the system.
Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for a n, the model for the sequence is
2
a n = –2n + n + 9.
36. −7, −2, 9, 26, 49, 78, …
SOLUTION: 2
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the system.
Therefore, a = –2, b = 1, and c = 9. Substituting these values in the equation for a n, the model for the sequence is
2
10-2 aArithmetic
Sequences and Series
= –2n + n + 9.
n
36. −7, −2, 9, 26, 49, 78, …
SOLUTION: 2
the system.
Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for a n, the model for the sequence is
2
a n = 3n – 4n – 6.
37. 6, −2, −12, −24, −38, −54, …
SOLUTION: 2
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the system.
Therefore, a = 3, b = –4, and c = –6. Substituting these values in the equation for a n, the model for the sequence is
2
10-2 aArithmetic
and Series
= 3n – 4n – Sequences
6.
n
37. 6, −2, −12, −24, −38, −54, …
SOLUTION: 2
the system.
Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for a n, the model for the sequence
2
is a n = –n – 5n + 12.
38. −3, 1, 13, 33, 61, 97, …
SOLUTION: 2
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the system.
Therefore, a = –1, b = –5, and c = 12. Substituting these values in the equation for a n, the model for the sequence
2
10-2 isArithmetic
a = –n – 5n + 12.
n
38. −3, 1, 13, 33, 61, 97, …
SOLUTION: 2
the system.
Next, enter the matrix into your graphing calculator, and use the rref( feature under the MATH menu to find the
solution.
Therefore, a = 4, b = –8, and c = 1. Substituting these values in the equation for a n, the model for the sequence is
2
a n = 4n – 8n + 1.
Find the indicated sum of each arithmetic series.
39. 26th partial sum of 3 + 15 + 27 + … + 303
SOLUTION: In this sequence, a 1 = 3 and d = 15 – 3 or 12. Find the sum of the series.
40. −28 + (−19) + (−10) + … + 242
SOLUTION: In this sequence, a 1 = –28, a n
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terms in the sequence n.
= 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of
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40. −28 + (−19) + (−10) + … + 242
SOLUTION: In this sequence, a 1 = –28, a n = 242, and d = –19 – (–28) or 9. Use the nth term formula to find the number of
Find the sum of the series.
41. 42nd partial sum of 120 + 114 + 108 + …
SOLUTION: In this sequence, a 1 = 120 and d = 114 – 120 or –6. Find the 42nd partial sum of the series.
42. 54th partial sum of 213 + 205 + 197 + …
SOLUTION: In this sequence, a 1 = 213 and d = 205 – 213 or –8. Find the 54th partial sum of the series.
43. −17 + 1 + 19 + … + 649
SOLUTION: eSolutions
In this sequence, a 1 = –17, a n = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of
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43. −17 + 1 + 19 + … + 649
SOLUTION: In this sequence, a 1 = –17, a n = 649, and d = 1 – (–17) or 18. Use the nth term formula to find the number of
44. 89 + 58 + 27 + … + (–562)
SOLUTION: In this sequence, a 1 = 89, a n = –562, and d = 58 – (89) or –31. Use the nth term formula to find the number of
45. RUNNING Refer to the beginning of the lesson.
a. Determine the number of miles Meg will run on her 12th day of training.
b. During which day of training will Meg reach her goal of 100 total miles?
SOLUTION: a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,
and so on. Therefore, a 1 = 1, a 2 = 1.25, and a 3 = 1.5. Find the common difference.
1.25 – 1 = 0.25
1.5 – 1.25 = 0.25
The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a 12.
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45. RUNNING Refer to the beginning of the lesson.
a. Determine the number of miles Meg will run on her 12th day of training.
b. During which day of training will Meg reach her goal of 100 total miles?
SOLUTION: a. From the lesson opener, Meg plans to run 1 mile the first day, 1.25 miles the second day, 1.5 miles the third day,
and so on. Therefore, a 1 = 1, a 2 = 1.25, and a 3 = 1.5. Find the common difference.
1.25 – 1 = 0.25
1.5 – 1.25 = 0.25
The common difference is 0.25. Use the formula for the nth term of an arithmetic sequence to find a 12.
th
So, Meg will run 3.75 miles on her 12 day of training.
b. Find the term that corresponds to S n = 100.
First, write an explicit formula using a 1 = 1 and d = 0.25.
Substitute S n = 100, a 1 = 1, and a n = 0.75 + 0.25n into the formula for the sum of a finite arithmetic series.
Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.
46. SOLUTION: eSolutions Manual - Powered by Cognero
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The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the
Because time cannot be negative, n = 25. Therefore, Meg will reach her goal of 100 total miles on day 25.
46. SOLUTION: The first term of this series is 5 and the last term is 43. The number of terms is equal to the upper bound minus the
lower bound plus one, which is 20 – 1 + 1 or 20. Therefore, a 1 = 5, a n = 43, and n = 20. Find the sum of the series.
47. SOLUTION: The first term of this series is 96 and the last term is –12. The number of terms is equal to the upper bound minus
the lower bound plus one, which is 28 – 1 + 1 or 28. Therefore, a 1 = 96, a n = –12, and n = 28. Find the sum of the
series.
48. SOLUTION: The first term of this series is –35 and the last term is –188. The number of terms is equal to the upper bound
minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a 1 = –89, a n = –188, and n = 12. Find the
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sum of the series.
minus the lower bound plus one, which is 18 – 7 + 1 or 12. Therefore, a 1 = –89, a n = –188, and n = 12. Find the
sum of the series.
49. SOLUTION: The first term of this series is 43 and the last term is 365. The number of terms is equal to the upper bound minus
the lower bound plus one, which is 52 – 6 + 1 or 47. Therefore, a 1 = 43, a n = 365, and n = 47. Find the sum of the
series.
Page 26
The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus
50. SOLUTION: The first term of this series is 63 and the last term is –42. The number of terms is equal to the upper bound minus
series.
51. SOLUTION: The first term of this series is 32 and the last term is 80. The number of terms is equal to the upper bound minus
the lower bound plus one, which is 13 – 1 + 1 or 13. Therefore, a 1 = 32, a n = 80, and n = 13. Find the sum of the
series.
Page 27
minus the lower bound plus one, which is 9 – 2 + 1 or 8. Therefore, a 1 = –42, a n = –147, and n = 8. Find the sum
of the series.
54. CONSTRUCTION A crew is tiling a hotel lobby with a trapezoidal mosaic pattern. The shorter base of the
trapezoid begins with a row of 8 tiles. Each row has two additional tiles until the 20th row. Determine the number
of tiles needed to create the mosaic design.
SOLUTION: The first row consists of 8 tiles and each row has two additional tiles until the 20th row, so a 1 = 8, d = 2, and n =
20. Find the sum of the series.
Therefore, 540 tiles are needed to create the mosaic design.
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Page 28
SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor
travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?
Therefore, 540 tiles are needed to create the mosaic design.
55. SNOWMOBILING A snowmobiling competitor travels 12 feet in the first second of a race. If the competitor
travels 1.5 additional feet each subsequent second, how many feet did the competitor travel in 64 seconds?
SOLUTION: The competitor travels 12 feet in the first second and 1.5 additional feet each subsequent second until the 64th
second. So, a 1 = 12, d = 1.5, and n = 64. Find the sum of the series.
Therefore, the competitor travels 3792 feet in 64 seconds.
56. FUNDRAISING Lalana organized a charity walk. In the first year, the walk generated $3000. She hopes to
increase this amount by $900 each year for the next several years. If her goal is met, in how many years will the
walk have generated a total of at least $65,000?
SOLUTION: The charity walk generated $3000 in the first year and Lalana hopes to increase the amount generated by $900
each year, so a 1 = 3000 and d = 900.
Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.
eSolutions
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Find a n if S n = 490, a 1 = –5, and n = 100.
SOLUTION: Page 29
Because time cannot be negative, it will take Lalana at least 10 years to generate a total of $65,000.
57. Find a n if S n = 490, a 1 = –5, and n = 100.
SOLUTION: Substitute S n = 490, a 1 = –5, and n = 100 into the formula for the sum of an arithmetic series.
58. If Sn = 51.7, n = 22, a n = –11.3, find a 1.
SOLUTION: Substitute S n = 51.7, n = 22, and a n = –11.3 into the formula for the sum of an arithmetic series.
59. Find n for –7 + (–5.5) + (–4) + … if S n = –14 and a n = 3.5.
SOLUTION: Substitute S n = –14, a n = 3.5, and a 1 = –7 into the formula for the sum of an arithmetic series.
60. Find a 1 if S n = 1287, n = 22, and d = 5.
SOLUTION: Substitute S n = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.
Page 30
60. Find a 1 if S n = 1287, n = 22, and d = 5.
SOLUTION: Substitute S n = 1287, n = 22, and d = 5 into the formula for the sum of an infinite arithmetic series.
61. If S 26 = 1456, and a 1 = –19, find d
SOLUTION: Substitute S 26 = 1456, n = 26, and a 1 = –19 into the formula for the sum of an arithmetic series.
62. If S 12 = 174, a 12 = 39, find d.
SOLUTION: Substitute a 12 = 39 and n = 12 into the formula for the nth term of an arithmetic sequence and solve for a 1.
Substitute S 12 = 174, a 1 = 39 – 11d, and n = 12 into the formula for the nth partial sum of an infinite arithmetic
series.
Write
each
arithmetic
series
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63. 6 + 12 + 18 + … + 66; n = 1
SOLUTION: in sigma notation. The lower bound is given.
Page 31
Write each arithmetic series in sigma notation. The lower bound is given.
63. 6 + 12 + 18 + … + 66; n = 1
12 – 6 = 6
18 – 12 = 6
Next, find n.
To find an explicit formula, substitute a 1 = 6 and d = 6 in the formula for the nth term of an arithmetic sequence.
Therefore, in sigma notation, the series 6 + 12 + 18 + . . . + 66 can be described as 64. –1 + 0 + 1 + … + 7; n = 1
0 – (–1) = 1
1– 0=1
Next, find n.
To find an explicit formula, substitute a 1 = –1 and d = 1 in the formula for the nth term of an arithmetic sequence.
Therefore, in sigma notation, the series – 1 + 0 + 1 + . . . 7 can be described as
.
65. 17 + 21 + 25 + … + 61; n = 4
SOLUTION: eSolutions Manual - Powered by Cognero
First, find the common difference.
21 – 17 = 4
25 – 21 = 4
Page 32
10-2 Therefore,
Arithmetic
Sequences
and
Series
in sigma
notation, the
series
– 1 + 0 + 1 + . . . 7 can be described as
.
65. 17 + 21 + 25 + … + 61; n = 4
21 – 17 = 4
25 – 21 = 4
Find a 1.
17 – 4 =13
13 – 4 = 9
9 – 4 = 5
Thus, a 1 = 5.
Next, find n.
Substitute a 1 = 5 and d = 4 in the formula for the nth term of an arithmetic sequence.
Therefore, in sigma notation, the series 17 + 21 + 25 + . . . + 61 can be described as
.
66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6
0 – 1 or –1
–1 – 0 or –1
Find a 1.
1 – (–1) = 2
2 – (–1) = 3
3 – (–1) = 4
4 – (–1) = 5
5 – (–1) = 6
Thus, a 1 = 6.
Next, find n. eSolutions Manual - Powered by Cognero
Page 33
10-2 Therefore,
Arithmetic
Sequences
and
Series
in sigma
notation, the
series
17 + 21 + 25 + . . . + 61 can be described as
.
66. 1 + 0 + (–1) + (–2) + … + (–13); n = 6
0 – 1 or –1
–1 – 0 or –1
Find a 1.
1 – (–1) = 2
2 – (–1) = 3
3 – (–1) = 4
4 – (–1) = 5
5 – (–1) = 6
Thus, a 1 = 6.
Next, find n. Substitute a 1 = 6 and d = –1 in the formula for the nth term of an arithmetic sequence.
Therefore, in sigma notation, the series 1 + 0 + (– 1) + (– 2) + . . . + (– 13) can be described as .
67. SOLUTION: First, find the common difference.
Find a 1.
a1 =
Next, n.
Page 34
10-2 Therefore,
Arithmetic
Sequences
and
Series
in sigma
notation, the
series
1 + 0 + (– 1) + (– 2) + . . . + (– 13) can be described as .
67. SOLUTION: First, find the common difference.
Find a 1.
a1 =
Next, n.
Substitute a 1 =
and d =
in the formula for the nth term of an arithmetic sequence.
Therefore, in sigma notation, the series
can be described as
.
68. 9.25 + 8.5 + 7.75 + … – 2; n = 1
8.5 – 9.25 = –0.75
7.75 – 8.5 = –0.75
Next, find n.
Page 35
Therefore, in sigma notation, the series
can be described as
. Sequences and Series
10-2 Arithmetic
68. 9.25 + 8.5 + 7.75 + … – 2; n = 1
8.5 – 9.25 = –0.75
7.75 – 8.5 = –0.75
Next, find n.
To find an explicit formula, substitute a 1 = 9.25 and d = –0.75 in the formula for the nth term of an arithmetic
sequence.
Therefore, in sigma notation, the series 9.25 + 8.5 + 7.75 + . . . + – 2 can be described as .
69. CONCERTS The seating in a concert auditorium is arranged as shown below.
a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown
in the first 3 rows continues for each successive row.
b. Find the total number of seats in the auditorium.
c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows.
How many seats are there in this auditorium?
SOLUTION: a. From the seating chart, a 1 = 24, a 2 = 29, a 3 = 34, and n = 35. Find the common difference.
29 – 24 = 5
34 – 29 = 5
Therefore, in sigma notation, the number of seats in the auditorium can be described as Page 36
.
in sigma
notation, the
series
9.25 + 8.5 + 7.75 + . . . + – 2 can be described as 10-2 Therefore,
Arithmetic
Sequences
and
Series
.
69. CONCERTS The seating in a concert auditorium is arranged as shown below.
a. Write a series in sigma notation to represent the number of seats in the auditorium, if the seating pattern shown
in the first 3 rows continues for each successive row.
b. Find the total number of seats in the auditorium.
c. Another auditorium has 32 rows with 18 seats in the first row and 4 more seats in each of the successive rows.
How many seats are there in this auditorium?
SOLUTION: a. From the seating chart, a 1 = 24, a 2 = 29, a 3 = 34, and n = 35. Find the common difference.
29 – 24 = 5
34 – 29 = 5
Therefore, in sigma notation, the number of seats in the auditorium can be described as .
b. From part a, a 1 = 24 and d = 5. Find S 35.
Therefore, there are 3815 seats in the auditorium.
c. For the second auditorium, n = 32, a 1 = 18, and d = 4. Find S 32.
Therefore, there are 2560 seats in the second auditorium.
Write a function that can be used to model the nth term of each sequence.
70. 2, 5,
8, 11,- Powered
14, 17, by
…Cognero
eSolutions
Manual
SOLUTION: Find the first differences.
Page 37
Therefore, there are 2560 seats in the second auditorium.
Write a function that can be used to model the nth term of each sequence.
70. 2, 5, 8, 11, 14, 17, …
The first differences are the same, so the nth term can be represented by a linear equation. Use the formula for
the nth term of an arithmetic sequence.
71. 8, 13, 20, 29, 40, 53, …
The first differences are not constant. Find the second differences.
The second differences are constant, so the sequence can be modeled by a quadratic equation of the form a n =
2
an + bn + c. Substitute values for a n and n into the equation.
the system.
Page 38
71. 8, 13, 20, 29, 40, 53, …
2
the system.
2
= n + 2n + 5.
72. 2, 2, 4, 8, 14, 22, …
Find the first differences.
Page 39
10-2 Arithmetic
2
= n + 2n + 5.
72. 2, 2, 4, 8, 14, 22, …
2
the system.
2
a n = n – 3n + 4.
73. 5, 31, 97, 221, 421, 715, …
Page 40
2
a n = n – 3n + 4.
73. 5, 31, 97, 221, 421, 715, …
The second differences are not constant. Find the third differences.
3
The third differences are constant, so the sequence can be modeled by a cubic equation of the form a n = an +
2
bn + cn + d. Substitute values for a n and n into the equation.
This yields a system of linear equations in four variables.
You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to
the system.
Page 41
Therefore, a = 3, b = 2, c = –1, and d = 1. Substituting these values in the equation for a n, the model for the
3
2
sequence is a n = 3n + 2n – n + 1.
74. –6, –8, –6, 6, 34, 84, …
3
The third differences are constant, so the sequence can be modeled by a cubic equation of the form a n = an +
2
bn + cn + d. Substitute values for a n and n into the equation.
This yields a system of linear equations in four variables.
You
can -use
a graphing
calculator
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the system.
to solve for a, b, c, and d. First, write the augmented matrix that corresponds
to 42
Page
You can use a graphing calculator to solve for a, b, c, and d. First, write the augmented matrix that corresponds to
the system.
Therefore, a = 1, b = –4, c = 3, and d = –6. Substituting these values in the equation for a n, the model for the
3
2
sequence is a n = n – 4n + 3n – 6.
75. 0, 23, 134, 447, 1124, 2375, …
The third differences are not constant. Find the fourth differences.
Page 43
The third differences are not constant. Find the fourth differences.
4
The fourth differences are constant, so the sequence can be modeled by a quartic equation of the form a n = an +
3
2
bn + cn + dn + e. Substitute values for a n and n into the equation.
This yields a system of linear equations in five variables.
You can use a graphing calculator to solve for a, b, c, d, and e. First, write the augmented matrix that corresponds
to the system.
Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for a n, the model for the
4
3
sequence is a n = 2n – n – 1.
Find each common difference.
76. Page 44
Therefore, a = 2, b = –1, c = 0, d = 0, and e = –1. Substituting these values in the equation for a n, the model for the
4
3
10-2 sequence
Arithmetic
Sequences
and Series
is a = 2n – n – 1.
n
Find each common difference.
76. SOLUTION: Find a 1 and a 100.
14 – 8 = 6
The expression in sigma notation is linear, so the common difference is the coefficient of the variable, 6.
77. SOLUTION: Find a 21 and a 65.
The expression in sigma notation is linear, so the common difference is the coefficient of the variable,
.
78. a 12 = 63, a 19 = 7
SOLUTION: eSolutions Manual - Powered by Cognero
a 19 is 7 terms from a 12.
Page 45
The expression in sigma notation is linear, so the common difference is the coefficient of the variable,
.
78. a 12 = 63, a 19 = 7
SOLUTION: a 19 is 7 terms from a 12.
79. a 8 = −4, a 27 =
SOLUTION: a 27 is 19 terms from a 8.
80. CALCULUS The area between the graph of a continuous function and the x-axis can be approximated using
2
sequences. Consider f (x) = x on the interval [1, 3].
a. Write the sequence xn formed when there are 5 arithmetic means between 1 and 3.
b. Write the sequence y n formed when y n = f (xn).
c. Write the sequence p n defined by d ⋅ y n.
d. The left-hand approximation of the area is given by
e . The right-hand approximation of the area is given by
. Find L6.
Find R6.
SOLUTION: a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,
where 3 is x7.
First, find the common difference using x7 = 3, x1 = 1, and n = 7.
Page 46
SOLUTION: a. A sequence from 1 to 3 with five arithmetic means will resemble 1, __?__ , __?__ , __?__ , __?__ , __?__ , 3,
10-2 where
Arithmetic
3 is x7. Sequences and Series
First, find the common difference using x7 = 3, x1 = 1, and n = 7.
Next, determine the arithmetic means by using d =
1+
= = = 2
2+
= = Therefore, a sequence with three arithmetic means between 1 and 3 is
b. Find each term in the sequence formed when y n = f (xn).
Page 47
Therefore, the sequence formed when y n = f (xn) is
c. Next, find each term in the sequence defined by d ⋅ y n.
Page 48
Therefore, the sequence defined by d ⋅ y n is
d.
e.
81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit
formula a n = 2 + 7(n − 1) for the sequence. Candace’s formula is a n = 7n − 5. Is either of them correct? Explain.
Manual
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Write
an -explicit
for
9– 2=7
16 – 9 = 7
the sequence. First, find the common difference.
Page 49
81. ERROR ANALYSIS Peter and Candace are given the arithmetic sequence 2, 9, 16, … . Peter wrote the explicit
formula a n = 2 + 7(n − 1) for the sequence. Candace’s formula is a n = 7n − 5. Is either of them correct? Explain.
SOLUTION: Write an explicit formula for the sequence. First, find the common difference.
9– 2=7
16 – 9 = 7
Next, substitute a 1 = 2 and d = 7 into the formula for the nth term of an arithmetic sequence.
Sample answer: Candace's formula a n = 7n − 5 is the simplified form of Peter's formula a n = 2 + 7(n − 1).
Therefore, both Candace and Peter are correct.
82. OPEN ENDED You have learned that the nth term of an arithmetic sequence can be modeled by a linear
function. Can the sequence of partial sums of an arithmetic series also be modeled by a linear function? If yes,
provide an example. If no, how can the sequence be modeled? Explain.
SOLUTION: Yes, the sequence of partial sums of an arithmetic series also be modeled by a linear function. Consider the sequence a 1, a 2, a 3, … . If the common difference of the sequence is 0, then the partial sums S n of
the corresponding series can be modeled by a linear function of the form S n = a 1n.
Otherwise, the partial sums can be modeled by a quadratic function of the form
where d
is the common difference of the original sequence.
83. CHALLENGE Prove that for an arithmetic sequence, a n = a k + (n – k)d for integers k in the domain of the
sequence.
Page 50
SOLUTION: Sample answer: The formula for the nth term of an arithmetic sequence is a n = a 1 + (n – 1)d. Similarly, a k = a 1 +
83. CHALLENGE Prove that for an arithmetic sequence, a n = a k + (n – k)d for integers k in the domain of the
sequence.
SOLUTION: Sample answer: The formula for the nth term of an arithmetic sequence is a n = a 1 + (n – 1)d. Similarly, a k = a 1 +
(k – 1)d. Solving the second equation for a 1 and substituting into the first yields:
REASONING Determine whether each statement is true or false for finite arithmetic series. Explain.
84. If you know the sum and d, you can solve for a 1.
SOLUTION: The is statement "If you know the sum and d, you can solve for a 1." is false. You must also know n.
Consider both formulas for the sum of a finite arithmetic series. and
. In the first formula, you must know Sn, n, and a n to find a 1. In
the second formula, you must know Sn, n, and a n and d to find a 1.
85. If you only know the first and last terms, then you can find the sum.
SOLUTION: The statement "If you only know the first and last terms, then you can find the sum." is false. You must also know
n.
Consider the two formulas for the sum of a finite arithmetic series. and
. For the first formula, in addition to the first and last terms, you
also need n. In the second formula, you need the first term, n, and d. 86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of
the series is positive.
SOLUTION: The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positive
or the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3
terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.
Consider the sequence 33, 24, 15, ...
Find the common difference.
24 – 33 = –9
15 – 24 = –9
Use the common difference of –9 to find the next three terms.
15 – 9 = 6
6 – 9 = –3
–3 – 9 = –12
The terms in the sequence are no all positive. Page 51
n.
Consider the two formulas for the sum of a finite arithmetic series. and
10-2 Arithmetic Sequences
and Series . For the first formula, in addition to the first and last terms, you
also need n. In the second formula, you need the first term, n, and d. 86. If the first three terms of a sequence are positive, then all of the terms of the sequence are positive or the sum of
the series is positive.
SOLUTION: The statement "If the first three terms of a sequence are positive, then all of the terms of the sequence are positive
or the sum of the series is positive." is false. If the common difference is negative, then it is possible for the first 3
terms to be positive, and the remaining terms to be negative. So, the sum could also be negative.
Consider the sequence 33, 24, 15, ...
Find the common difference.
24 – 33 = –9
15 – 24 = –9
Use the common difference of –9 to find the next three terms.
15 – 9 = 6
6 – 9 = –3
–3 – 9 = –12
The terms in the sequence are no all positive. Find the sum of the first 10 terms in the sequence. The sum is negative. 87. CHALLENGE Consider the arithmetic sequence of odd natural numbers.
a. Find S 7 and S 9.
b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series.
c. Write an algebraic proof verifying the conjecture that you made in part b.
SOLUTION: a. The arithmetic sequence of odd natural numbers is a 1 = 1, a 2 = 3, a 3 = 5, a 5 = 7, a 6 = 9, a 7 = 11, a 8 = 13, a 9 =
15… . The common difference is 2.
Find S 7.
Find
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The sum is negative. 87. CHALLENGE Consider the arithmetic sequence of odd natural numbers.
a. Find S 7 and S 9.
b. Make a conjecture about the pattern that results from the sums of the corresponding arithmetic series.
c. Write an algebraic proof verifying the conjecture that you made in part b.
SOLUTION: a. The arithmetic sequence of odd natural numbers is a 1 = 1, a 2 = 3, a 3 = 5, a 5 = 7, a 6 = 9, a 7 = 11, a 8 = 13, a 9 =
15… . The common difference is 2.
Find S 7.
Find S 9.
b. Sample answer: Since the sum of the first 7 odd natural numbers is 49 and the sum of the first 9 is 81, the sum
2
of the first n terms of the sequence of odd natural numbers appears to be n .
c. Write an explicit formula for the arithmetic sequence of odd natural numbers.
Substitute a n = 2n – 1 and a 1 = 1 into the formula for the sum of a finite arithmetic series.
88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.
SOLUTION: Sample
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- Powered by Cognero
Page 53
88. WRITING IN MATH Explain why the arithmetic series 25 + 20 + 15 + … does not have a sum.
SOLUTION: Sample answer:
The partial sum S n is
which approaches –
as n
. Therefore, the sum of the series 25 + 20
+ 15 + … cannot be calculated.
Find the next four terms of each sequence.
89. 12, 16, 20, …
SOLUTION: These terms appear to increase by 4. Check.
16 – 12 = 4
20 – 16 = 4
The next four terms are 20 + 4 = 24
24 + 4 = 28
28 + 4 = 32
32 + 4 = 36.
90. 3, 1, −1, …
SOLUTION: The terms appear to decrease by 2. 1 – 3 = –2
–1 – 1 = –2
The next four terms are :
–1 –2 = –3
–3 –2 = –5
–5 –2 = –7
–7 –2 = –9.
91. 31, 24, 17, …
SOLUTION: These terms appear to decrease by 7. Check.
24Manual
– 31 =- Powered
–7
eSolutions
by Cognero
17 – 24 = –7
The next four terms are:
Page 54
The next four terms are :
–1 –2 = –3
–3 –2 = –5
–2 = –7
10-2 –5
Arithmetic
Sequences
–7 –2 = –9.
and Series
91. 31, 24, 17, …
SOLUTION: These terms appear to decrease by 7. Check.
24 – 31 = –7
17 – 24 = –7
The next four terms are:
17 – 7 = 10
10 – 7 = 3
3 – 7 = –4
–4 – 7 = –11 .
Find each product or quotient and express it in rectangular form.
92. SOLUTION: Use the Product Formula to find the product in polar form.
Now find the rectangular form of the product.
The polar form of the product is
. The rectangular form of the quotient is
.
93. SOLUTION: Use the Quotient Formula to find the quotient in polar form.
Page 55
polar form of the product is
10-2 The
Arithmetic
.
93. SOLUTION: Use the Quotient Formula to find the quotient in polar form.
The polar form of the quotient is
.
Find the dot product of u and v. Then determine if u and v are orthogonal.
,v=
94. u =
SOLUTION: , u and v are not orthogonal.
Since
95. u =
,v=
SOLUTION: , u and v are not orthogonal.
Since
96. u =
,v=
SOLUTION: Since
, u and v are not orthogonal.
Find the direction angle of each vector to the nearest tenth of a degree.
97. – i – 3j
SOLUTION: Page 56
Since
, u and v are not orthogonal.
Find the direction angle of each vector to the nearest tenth of a degree.
97. – i – 3j
SOLUTION: Since the vector lies in Quadrant III, θ = 180° + 71.6° or about 251.6°.
98. SOLUTION: Since the vector lies in Quadrant II, θ = 180° + (−29.1)° or about 150.9°.
Page 57
99. SOLUTION: Since the vector lies in Quadrant II, θ = 180° + (−45)° or 135 °.
100. MANUFACTURING A cam in a punch press is shaped like an ellipse with the equation
+ = 1. The camshaft goes through the focus on the positive axis.
a. Graph a model of the cam.
b. Find an equation that translates the model so that the camshaft is at the origin.
c. Find the equation of the model in part b when the cam is rotated to an upright position.
SOLUTION: a.
b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented
Page 58
ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.
b. To find the current location of the cam, find the location of the right focus. The foci of a horizontally oriented
ellipse are given by (h ± c, k). From the equation, h = 0 and k = 0. Find c.
The right focus is located at (
, 0), so the graph of the focus to be at the origin. Therefore, the translated equation is
needs to be shifted
units to the left for .
c. For the cam to be rotated to an upright position, it needs to be rotated counterclockwise through an angle of 90º.
First, find the equations for x and y.
Substitute these values into the original equation.
Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by
.
101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then
sketch the graphs of f and g.
SOLUTION: This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the
right and expanded vertically.
Page 59
Therefore, the equation of the model in part b when the cam is rotated to an upright position is given by
10-2 Arithmetic Sequences
and Series
.
101. Use the graph of f (x) = ln x to describe the transformation that results in the graph of g(x) = 3 ln (x − 1). Then
sketch the graphs of f and g.
SOLUTION: This function is of the form g(x) = 3f (x – 1). Therefore, the graph of g(x) is the graph of f (x) shifted 1 unit to the
right and expanded vertically.
102. SAT/ACT What is the units digit of 336?
A0
B1
C3
D7
E9
SOLUTION: n
Find 3 for whole number values of n, where 1 ≤ n ≤ 8.
1
5
2
6
3
7
4
Notice that there is a pattern emerging: 3 = 3 and 3 = 243 , 3 = 9 and 3 = 729 , 3 = 27 and 3 = 2187 , and 3 =
8
36
81 and 3 = 6561 . From this pattern, you can conclude that 3 will also have a units digit of 1. Therefore, the
correct answer is B.
103. Using the table, which formula can be used to determine the nth term of the sequence?
F a n = 6n
G an = n + 5
H a n = 2n + 1
J aManual
2
eSolutions
Powered
by Cognero
n = 4n- +
SOLUTION: Page 60
1
5
2
6
3
7
4
Notice that there is a pattern emerging: 3 = 3 and 3 = 243 , 3 = 9 and 3 = 729 , 3 = 27 and 3 = 2187 , and 3 =
8
36
1 and 3 = 6561Sequences
. From this pattern,
can conclude that 3 will also have a units digit of 1. Therefore, the
10-2 8Arithmetic
and you
Series
correct answer is B.
103. Using the table, which formula can be used to determine the nth term of the sequence?
F a n = 6n
G an = n + 5
H a n = 2n + 1
J a n = 4n + 2
SOLUTION: From the table, a 1 = 6, a 2 = 10, a 3 = 14, and a 4 = 18. The common difference is 10 – 6 or 4.
Write an explicit formula for the nth term of the sequence.
Therefore, the correct answer is J.
104. REVIEW If a 1 = 3, a 2 = 5, and a n = a n − 2 + 3n, find a 10.
A 59
B 75
C 89
D 125
SOLUTION: Substitute n = 10 into a n = a n − 2 + 3n.
To find a 10, you must find a 8. However, to find a 8, you must first find a 3, a 4, …, a 6.
Page 61
Therefore, the correct answer is J.
104. REVIEW If a 1 = 3, a 2 = 5, and a n = a n − 2 + 3n, find a 10.
A 59
B 75
C 89
D 125
SOLUTION: Substitute n = 10 into a n = a n − 2 + 3n.
To find a 10, you must find a 8. However, to find a 8, you must first find a 3, a 4, …, a 6.
Now that a 6 is known, you can find a 8.
Substitute a 8 = 59 into the equation for a 10.
Therefore, the correct answer is C.
105. REVIEW Which of the sequences shown below is convergent?
Page 62
Therefore, the correct answer is C.
105. REVIEW Which of the sequences shown below is convergent?
SOLUTION: A sequence is convergent if the terms approach a unique number. After examining each graph, it appears that only
the sequence that approaches a unique number is the one shown in the graph for choice J. Therefore, the correct
answer is J.
Page 63

10-2 Arithmetic Sequences and Series

Transcription

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