10-2 Arithmetic Sequences and Series

10-2 Arithmetic Sequences and Series Find the indicated term of each arithmetic sequence.

1. a 1 = 14, d = 9, n = 11

SOLUTION:

2. a 18 for 12, 25, 38, …

SOLUTION: 25 – 12 = 13 38 – 25 = 13

The common difference d is 13.

a 1 = 12

Write an equation for the nth term of each arithmetic sequence.

3. 13, 19, 25, …

SOLUTION: 19 – 13 = 6 25 – 19 = 6


a 1 = 13

4. a 5 Manual = –12,-dPowered = –4 by Cognero eSolutions

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10-2 Arithmetic Sequences and Series 4. a 5 = –12, d = –4

SOLUTION:

Use the value of a 1 to find the nth term.

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.

6.

SOLUTION: Here a 1 = –4 and a 5 = 8.

Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5.

Manual - Powered by Cognero eSolutions Find the sum of each arithmetic series.

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Therefore, the missing numbers are (6 + 9) or 15, (15 + 9) or 24, (24 + 9) or 33. 10-2 Arithmetic Sequences and Series 6.


Therefore, the missing numbers are (–4 + 3) or −1, (–1 + 3) or 2, (2 + 3) or 5.

Find the sum of each arithmetic series.

7. the first 50 natural numbers

SOLUTION:

8. 4 + 8 + 12 + … + 200

SOLUTION: 8– 4=4 12 – 8 = 4

The common difference is 4.

Find the sum of the series.

eSolutions Manual - Powered by Cognero

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10-2 Arithmetic Sequences and Series 8. 4 + 8 + 12 + … + 200

SOLUTION: 8– 4=4 12 – 8 = 4



9. a 1 = 12, a n = 188, d = 4

SOLUTION:


10. a n = 145, d = 5, n = 21

SOLUTION: eSolutions Manual - Powered by Cognero

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10-2 Arithmetic Sequences and Series

10. a n = 145, d = 5, n = 21

SOLUTION:


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.

12. n = 18, a n = 112, S n = 1098

SOLUTION:


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Therefore, the first three terms are 8, (8 + 4) or 12, (12 + 4) or 16. 10-2 Arithmetic Sequences and Series 12. n = 18, a n = 112, S n = 1098

SOLUTION:

Therefore, the first three terms are 10, (10 + 6) or 16, (16 + 6) or 22.

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.

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Therefore, the first three and termsSeries are 10, (10 + 6) or 16, (16 + 6) or 22. 10-2 Arithmetic Sequences

13. MULTIPLE CHOICE Find

A 45

B 78

C 342

D 410


Find the sum.

Option C is the correct answer.

Find the indicated term of each arithmetic sequence.

14. a 1 = –18, d = 12, n = 16

SOLUTION:

15. a 1 = –12, n = 66, d = 4

SOLUTION:


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10-2 Arithmetic Sequences and Series 15. a 1 = –12, n = 66, d = 4

SOLUTION:

16. a 1 = 9, n = 24, d = –6

SOLUTION:

17. a 15 for –5, –12, –19, …

SOLUTION: –12 – (–5) = –7 –19 – (–12) = –7

The common difference d is –7.

a 1 = –5

18. a 10 for –1, 1, 3, …

SOLUTION: 1 – (–1) = 2 3– 1=2


a 1 = –1


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18. a 10 for –1, 1, 3, …

SOLUTION: 1 – (–1) = 2 3– 1=2


a 1 = –1

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


20. 24, 35, 46, …

SOLUTION: 35 – 24 = 11 46 – 35 = 11


a 1 = 24


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20. 24, 35, 46, …

SOLUTION: 35 – 24 = 11 46 – 35 = 11


a 1 = 24

21. 31, 17, 3, …

SOLUTION: 17 – 31 = –14 3 – 17 = –14


a 1 = 31

22. a 9 = 45, d = –3

SOLUTION:

Use the value of a 1 to find the nth term. eSolutions Manual - Powered by Cognero

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22. a 9 = 45, d = –3

SOLUTION:


23. a 7 = 21, d = 5

SOLUTION:


24. a 4 = 12, d = 0.25

SOLUTION:



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24. a 4 = 12, d = 0.25

SOLUTION:


25. a 5 = 1.5, d = 4.5

SOLUTION:


26. 9, 2, –5, …

SOLUTION: 2 – 9 = –7 –5 – 2 = –7


a1 = 9


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26. 9, 2, –5, …

SOLUTION: 2 – 9 = –7 –5 – 2 = –7


a1 = 9

27. a 6 = 22, d = 9

SOLUTION:


28. a 8 = –8, d = –2

SOLUTION:



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28. a 8 = –8, d = –2

SOLUTION:


29.

SOLUTION:


30. –12, –17, –22, …

SOLUTION: –17 – (–12) = –5 –22 – (–17) = –5

eSolutions - Powered by Cognero TheManual common difference d is

a 1 = –12

–5.

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30. –12, –17, –22, …

SOLUTION: –17 – (–12) = –5 –22 – (–17) = –5


a 1 = –12

31.

SOLUTION:


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.

a. Write an equation to represent the nth term of the 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. eSolutions Manual - Powered by Cognero

SOLUTION:

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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.


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.

b. Substitute 187 for a n and solve for n.

th

Therefore José’s average will be 187 pins per game in the 9 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.

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.

34.

]

eSolutions Manual - Powered by Cognero SOLUTION:

Here a 1 = –6 and a 6 = 49.

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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 34.

]


Therefore, the missing numbers are (–6 + 11) or 5, (5 + 11) or 16, (16 + 11) or 27, and (27 + 11) or 38.

35.


Therefore, the missing numbers are (–28 + 7) or –21, (–21 + 7) or –14, (–14 + 7) or –7, and (–7 + 7) or 0.

36.


Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48.

37.

SOLUTION: eSolutions Manual Here a 1 =- Powered –12 andbyaCognero 7 = –66.

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Therefore, the missing numbers are (84 – 9) or 75, (75 – 9) or 66, (66 – 9) or 57, and (57 – 9) or 48. 10-2 Arithmetic Sequences and Series

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.

38.


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.


39. the first 100 even natural numbers


Find the sum.


40. the first 200 odd natural numbers

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Therefore, the missing numbers are (182 – 13) or 169, (169 – 13) or 156, (156 – 13) or 143, (143 – 13) or 130, and (130 – 13) orSequences 117. 10-2 Arithmetic and Series




Find the sum.



Find the sum.



n = 100

Find the sum.



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n = 100

Find the sum.



n = 300

Find the sum.

43. –18 + (–15) + (–12) + … + 66

SOLUTION: –15 – (–18) = 3 –12 – (–15) = 3


Find the value of n.


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43. –18 + (–15) + (–12) + … + 66

SOLUTION: –15 – (–18) = 3 –12 – (–15) = 3



Find the sum.

44. –24 + (–18) + (–12) + … + 72

SOLUTION: –18 – (–24) = 6 –12 – (–18) = 6



eSolutions Manual - Powered by Cognero Find the sum.

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44. –24 + (–18) + (–12) + … + 72

SOLUTION: –18 – (–24) = 6 –12 – (–18) = 6



Find the sum.

45. a 1 = –16, d = 6, n = 24

SOLUTION:

46. n = 19, a n = 154, d = 8

SOLUTION: Find the value of a 1.


Find the sum.

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46. n = 19, a n = 154, d = 8


Find the sum.

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.


48. n = 32, a n = –86, S n = 224



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A cash prizesSequences totaled $4400 the eleven week contest. 10-2 Arithmetic andfor Series


48. n = 32, a n = –86, S n = 224


Find the value of d.

So, the sequence is 100, 94, 88, …

49. a 1 = 48, a n = 180, S n = 1368

SOLUTION: Find the value of n.




50. a = 3, a = 66, S = 759

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So, the sequence is 100, 94, 88, … 10-2 Arithmetic Sequences and Series 49. a 1 = 48, a n = 180, S n = 1368




50. a 1 = 3, a n = 66, S n = 759



Therefore, the first three terms are 3, 6 and 9.

51. n = 28, a n = 228, S n = 2982



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Therefore, the first threeand terms are 3, 6 and 9. 10-2 Arithmetic Sequences Series

51. n = 28, a n = 228, S n = 2982



Therefore, the first three terms are –15, –6 and 3.

52. a 1 = –72, a n = 453, S n = 6858



Therefore, the first three terms are –72, –57 and –42.


53. n = 30, a n = 362, S n = 4770

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Therefore, the first threeand terms are –15, –6 and 3. 10-2 Arithmetic Sequences Series

52. a 1 = –72, a n = 453, S n = 6858




53. n = 30, a n = 362, S n = 4770




54. a = 19, n = 44, S = 9350

1 Manual - Powerednby Cognero eSolutions

SOLUTION:

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Therefore, the first threeand terms are –44, –30 and –16. 10-2 Arithmetic Sequences 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





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56. PRIZES A radio station is offering a total of $8500 in prizes over ten hours. Each hour, the prize will increase by

Therefore, the first three and termsSeries are 19, 28 and 37. 10-2 Arithmetic Sequences

55. a 1 = –33, n = 36, S n = 6372




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.





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Therefore, the first threeand terms are –33, –21 and –9. 10-2 Arithmetic Sequences 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.

SOLUTION: Given n = 10, d = 100 and S 10 = 8500.




57.


Find the sum.

Therefore, eSolutions Manual - Powered by Cognero

. Page 30



57.


Find the sum.

Therefore,

.

58.


Find the sum.

Therefore,

.


59.

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Therefore,

.


58.


Find the sum.

Therefore,

.

59.


Find the sum.

Therefore,

.

60. eSolutions Manual - Powered by Cognero

SOLUTION:

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Therefore, . 10-2 Arithmetic Sequences and Series

60.


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.

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?

SOLUTION: Given a 1 = 16, n = 10.



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She pays $2250. 10-2 Arithmetic Sequences and Series

62. GRAVITY When an object is in free fall and air resistance is ignored, it falls 16 feet in the first second, an additional 48 feet during the next second, and 80 feet during the third second. How many total feet will the object fall in 10 seconds?

SOLUTION: Given a 1 = 16, n = 10.


Find the sum.

The object will fall 1600 ft in 10 seconds.

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.


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.


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10-2 Arithmetic Sequences and Series 64. The eleventh term of the sequence is 78. The common difference is –9.

SOLUTION: Given a 11 = 78, d = –9 and n = 11.



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.




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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.



66. The 25th term of the sequence is 121. The 80th term is 506.

SOLUTION: Given a 25 = 121 and a 80 = 506.

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.



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66. The 25th term of the sequence is 121. The 80th term is 506.

SOLUTION: Given a 25 = 121 and a 80 = 506.

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.



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. eSolutions Manual - Powered by Cognero

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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

.

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?

SOLUTION: Given a 1 = 200, d = 24 and a n = 512.

Substitute the values of a 1, a n and d and solve for n.


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c. No; there is no whole number n for which 10-2 Arithmetic Sequences and Series

.

68. PERFORMANCE A certain company pays its employees according to their performance. Belinda is paid a flat rate of $200 per week plus $24 for every unit she completes. If she earned $512 in one week, how many units did she complete?



14th term in the sequence is 512.

Therefore, she completed (14 – 1) or 13 units.

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?



So he will have a salary of $100,000 per year after the 19th year.

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.


b. If the pattern continues, during which week will she be running 10 miles per day?


SOLUTION: a. Given a 1 = 3 and d = 0.5.


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So he will have a salary of $100,000 10-2 Arithmetic Sequences and Series per year after the 19th year.

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.


b. If the pattern continues, during which week will she be running 10 miles per day?


SOLUTION: a. Given a 1 = 3 and d = 0.5.

Find the nth term.

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.

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).

2

c. GRAPHICAL Graph f (x) = x + 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?

f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x2 + 8x.

SOLUTION: eSolutions a. Manual - Powered by Cognero

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sum of arithmetic series?

f. ALGEBRAIC Find the arithmetic series that relates to g(x) = x2 + 8x. 10-2 Arithmetic Sequences and Series SOLUTION: a.

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.

72.


SOLUTION: There are x – 3 + 1 or x – 2 terms, so n = x – 2.

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shares the same range.

10-2 Arithmetic Sequences and Series f. Find the value of x.

72.


Find the sum.

Equate the sum with the given value and solve for x.

The value of x should be positive. Therefore, x = 18.

73.


Manual - Powered by Cognero eSolutions Find the sum.

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The value of x should be positive. Therefore, x = 18. 10-2 Arithmetic Sequences and Series

73.


Find the sum.

Equate the sum with the given value and solve for x.

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.


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The value ofSequences x should be and positive. Therefore, x = 16. 10-2 Arithmetic Series

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.

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.


Find the value of c in terms of a and b.


Page 44

SOLUTION: Sample answer: Eric; Juana 10-2 Arithmetic Sequences andmissed Seriesthe 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.


Find the value of c in terms of a and b.

76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?

SOLUTION: The three arithmetic means between a and b are

.

The average of the arithmetic means is

.

Therefore, . The term b can be written as a + 4d.

The average of a and b is

.

We know that

.

eSolutions Manual - Powered by Cognero Therefore, the average of a

and b is 16.

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76. CHALLENGE There are three arithmetic means between a and b in an arithmetic sequence. The average of the arithmetic means is 16. What is the average of a and b?

SOLUTION: The three arithmetic means between a and b are

.

The average of the arithmetic means is

.

Therefore, . The term b can be written as a + 4d.

The average of a and b is

.

We know that

.

Therefore, the average of a and b is 16.

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.

78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324. eSolutions Manual - Powered by Cognero

SOLUTION:

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78. OPEN ENDED Write an arithmetic series with 8 terms and a sum of 324.

SOLUTION: Sample answer: 9 + 18 + 27 + … + 72

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.

80. PROOF Prove the formula for the nth term of an arithmetic sequence.

SOLUTION: Sample answer: Let a n = the nth term of the sequence and d = the common difference a 2 = a 1 + d Definition of the second term of an arithmetic sequence a 3 = a 2 + d Definition of the third term of an arithmetic sequence a 3 = (a 1 + d ) + d Substitution a 3 = a 1 + 2d Associative Property of Addition a 3 = a 1 + (3 – 1)d 3 – 1 = 2 a n = a 1 + (n – 1)d n = 3

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.


82. PROOF Derive the Alternate Sum Formula using the General Sum Formula.

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Simplify. 10-2 Arithmetic Sequences and Series

82. PROOF Derive the Alternate Sum Formula using the General Sum Formula.

SOLUTION: General sum formula

a n = a 1 + (n – 1)d Formula for nth term

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. 84. The area of a triangle is

and the height is q + 4. Which expression best describes the triangle’s length?

F (q + 1)


G (q + 2)

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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. 84. The area of a triangle is

and the height is q + 4. Which expression best describes the triangle’s length?

F (q + 1)

G (q + 2)

H (q – 3)

J (q – 4)

SOLUTION: The area of the triangle is

.

The length of the triangle is

Option J is the correct answer.

85. The expression

is equivalent to

A

B

C

D

SOLUTION:

Option A is the correct answer.

Trevor 86. SHORT eSolutions Manual -RESPONSE Powered by Cognero

can type a 200-word essay in 6 hours. Minya can type the same essay in

If they work together, how many hours will it take them to type the essay?

hours. Page 49

Option A is the correct answer. 10-2 Arithmetic Sequences and Series

86. SHORT RESPONSE Trevor can type a 200-word essay in 6 hours. Minya can type the same essay in

hours.

If they work together, how many hours will it take them to type the essay?

SOLUTION: Trevor can type

words in an hour.

Minya can type

words in an hour.

They type

words together in an hour.

They will type the essay in

hours if they work together.

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.

88.

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.

Manual - Powered by Cognero eSolutions Solve each system of inequalities by graphing.

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SOLUTION: Since there isSequences a common and difference 10-2 Arithmetic Seriesbetween 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.

90.

SOLUTION:

91.

SOLUTION:

92.

SOLUTION:


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92.

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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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.

Manual - Powered by Cognero eSolutions 94.

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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. Sequences and Series 10-2 Arithmetic

Graph each function. State the domain and range.

94.

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 0 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) > 0}.

95.




TheManual value- Powered of the f (x) tends to eSolutions by Cognero

3 as x tends to –∞.


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Therefore, the range of the function is R ={f (x) | f (x) > 0}. 10-2 Arithmetic Sequences and Series 95.




The value of the f (x) tends to 3 as x tends to –∞.


Therefore, the range of the function is R ={f (x) | f (x) > 3}.

96.




The value of the f (x) tends to ∞ as x tends to –∞.

TheManual value- Powered of the f (x) tends to eSolutions by Cognero

–1 as x tends to ∞.

Therefore, the range of the function is R ={f (x) | f (x) > –1}.

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Therefore, the range of the function 10-2 Arithmetic Sequences and Seriesis R ={f (x) | f (x) > 3}.

96.




The value of the f (x) tends to ∞ as x tends to –∞.

The value of the f (x) tends to –1 as x tends to ∞.

Therefore, the range of the function is R ={f (x) | f (x) > –1}.

Solve each equation. Round to the nearest ten-thousandth.

x

97. 5 = 52

SOLUTION:

3p

98. 4

= 10

SOLUTION:


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10-2 Arithmetic Sequences and Series 3p

98. 4

= 10

SOLUTION:

n+2

99. 3

= 14.5

SOLUTION:

d –4

100. 16

3–d

=3

SOLUTION:


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10-2 Arithmetic Sequences and Series

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