ms. wilson's math classes -...

Determine the common ratio, and find the next three terms of each geometric sequence.

1.

SOLUTION: First, find the common ratio.

= –2

The common ratio is –2. Multiply the third term by –2 to find the fourth term, and so on. –1(–2) = 2 2(–2) = –4 –4(–2) = 8 Therefore, the next three terms are 2, –4, and 8.

2.

The common ratio is . Multiply the third term

by to find the fourth term, and so on.

Therefore, the next three terms are – , ,

and – .

3. 0.5, 0.75, 1.125, …

SOLUTION: First, find the common ratio. 0.75 ÷ 0.5 = 1.5 1.125 ÷ 0.75 = 1.5 The common ratio is 1.5. Multiply the third term by 1.5 to find the fourth term, and so on. 1.125(1.5) = 1.6875 1.6875(1.5) = 2.53125 2.53125(1.5) = 3.796875 Therefore, the next three terms are 1.6875, 2.53125, and 3.796875.

4. 8, 20, 50, …

SOLUTION: First, find the common ratio. 20 ÷ 8 or 2.5 50 ÷ 20 or 2.5 The common ratio is 2.5. Multiply the third term by 2.5 to find the fourth term, and so on. 50(2.5) = 125 125(2.5) = 312.5 312.5(2.5) = 781.25 Therefore, the next three terms are 125, 312.5, and 781.25.

5. 2x, 10x, 50x, …

SOLUTION: First, find the common ratio. 10x ÷ 2x = 5 50x ÷ 10x = 5 The common ratio is 5. Multiply the third term by 5 to find the fourth term, and so on. 5(50x) = 250x 5(250x) =1250x 5(1250x) =6250x Therefore, the next three terms are 250x, 1250x, and 6250x.

6. 64x, 16x, 4x, …

16x ÷ 64x =

4x ÷ 16x =

The common ratio is . Multiply the third term by

to find the fourth term, and so on.

Therefore, the next three terms are x, x, and

x.

7. x + 5, 3x + 15, 9x + 45, …

The common ratio is 3. Multiply the third term by 3 to find the fourth term, and so on. 3(9x + 45) = 27x +135 3(27x +135) = 81x + 405 3(81x + 405) = 243x + 1215 Therefore, the next three terms are 27x +135, 81x + 405, and 243x + 1215.

8. –9 – y , 27 + 3y , –81 – 9y , …

The common ratio is –3. Multiply the third term by –3 to find the fourth term, and so on. –3(–81 – 9y) = 243 + 27y –3(243 + 27y) = –729 – 81y –3(–729 – 81y) = 2187 + 243y Therefore, the next three terms are 243 + 27y , –729 – 81y , and 2187 + 243y .

9. GEOMETRY Consider a sequence of circles withdiameters that form a geometric sequence: d1, d2,

d3, d4, d5.

a. Show that the sequence of circumferences of the circles is also geometric. Identify r. b. Show that the sequence of areas of the circles isalso geometric. Identify the common ratio.

SOLUTION: a. Sample answer: The circumference of a circle isgiven by C = πd. So, the sequence of circumferences of the circles is πd1, πd2, πd3, πd4,

πd5.

Find the common ratio.

b. Sample answer: The area of a circle is given by

C = πr2 or So, the sequence of areas

of the circles is

Write an explicit formula and a recursive formula for finding the nth term of each geometric sequence.

10. 36, 12, 4, …

12 ÷ 36 =

4 ÷ 12 =

For an explicit formula, substitute a1 = 36 and r =

in the nth term formula.

For a recursive formula, state the first term a1.

Then indicate that the next term is the product of

the first term an – 1 and r.

a1 = 36, an =

11. 64, 16, 4, …

16 ÷ 64 =

4 ÷ 16 =

a1 = 64, an =

12. −2, 10, −50, …

SOLUTION: First, find the common ratio. 10 ÷ –2 = –5 –50 ÷ 10 = –5 For an explicit formula, substitute a1 = –2 and r = –

5 in the nth term formula.

a1 = –2, an =

13. 4, −12, 36, …

SOLUTION: First, find the common ratio. –12 ÷ 4 = –3 36 ÷ –12 = –3 For an explicit formula, substitute a1 = 4 and r = –3

a1 = 4, an =

14. 4, 8, 16, …

SOLUTION: First, find the common ratio. 8 ÷ 4 = 2 16 ÷ 8 = 2 For an explicit formula, substitute a1 = 4 and r = 2

a1 = 4, an =

15. 20, 30, 45, …

SOLUTION: First, find the common ratio. 30 ÷ 20 = 1.5 45 ÷ 30 = 1.5 For an explicit formula, substitute a1 = 20 and r =

1.5 in the nth term formula.

a1 = 20, an =

16. 15, 5, , …

5 ÷ 15 =

÷ 5 =

a1 = 15, an =

17. , , , …

÷ = 2

For an explicit formula, substitute a1 = and r =

a1 = , an =

18. CHAIN E-MAIL Melina receives a chain e-mail that she forwards to 7 of her friends. Each of her friends forwards it to 7 of their friends.

a. Write an explicit formula for the pattern. b. How many will receive the e-mail after 6 forwards?

SOLUTION: Melina receives a chain e-mail, forwards it to 7 friends, and each friend forwards it to 7 friends.

Therefore, a1 = 1, a2 = 7, and a3 = 49. The

common ratio is 7. For an explicit formula, substitute a1 = 1 and r = 7

b. Use the explicit formula you found in part a to find a6.

Therefore, after 6 forwards 16,807 people will havereceived the e-mail.

19. BIOLOGY A certain bacteria divides every 15 minutes to produce two complete bacteria. a. If an initial colony contains a population of b0

bacteria, write an equation that will determine the

number of bacteria bt present after t hours.

b. Suppose a Petri dish contains 12 bacteria. Use the equation found in part a to determine the number of bacteria present 4 hours later.

SOLUTION: a. Initially, there is 1 bacterium. After 15 minutes, there will be 2 bacteria, after 30 minutes there will be 4 bacteria, after 45 minutes there will be 8 bacteria, and after 1 hour there will be 16 bacteria.

So, in terms of hours, b0 = 1 and b1 = 16. Find the

common ratio.

16 ÷ 1 = 16 Write an explicit formula using r = 16.

b. Substitute b0 = 12 and t = 4 into the equation you

found in part a.

Find the specified term for each geometric sequence or sequence with the given characteristics.

20. a9 for 60, 30, 15, …

30 ÷ 60 =

15 ÷ 30 =

Use the formula for the nth term of a geometric

sequence to find a9.

21. a4 for 7, 14, 28, …

SOLUTION: First, find the common ratio. 14 ÷ 7 = 2 28 ÷ 14 = 2 Use the formula for the nth term of a geometric

22. a5 for 3, 1, , …

1 ÷ 3 =

÷ 1 =

23. a6 for 540, 90, 15, …

90 ÷ 540 =

15 ÷ 90 =

24. a7 if a3 = 24 and r = 0.5

SOLUTION:

Use the values of a3 and r to find a2.

Next, use the values of a2 and r to find a1.

Another method would be to consider that the 7th term is 4 terms from the 3rd term. Therefore,

multiply the 3rd term by r4.

25. a6 if a3 = 32 and r = –0.5

SOLUTION:

26. a6 if a1 = 16,807 and r =

SOLUTION: Use the formula for the nth term of a geometric

27. a8 if a1 = 4096 and r =

28. ACCOUNTING Julian Rockman is an accountantfor a small company. On January 1, 2009, the company purchased $50,000 worth of computers, printers, scanners, and hardware. Because this equipment is a company asset, Mr. Rockman needsto determine how much the computer equipment is presently worth. He estimates that the computer equipment depreciates at a rate of 45% per year. What value should Mr. Rockman assign the equipment in his 2014 year-end accounting report?

SOLUTION:

The equipment is originally worth $50,000, so a1 =

50,000. Because the equipment depreciates at a rate of 45% per year, the value of the equipment ona given year will be 100% – 45% or 55% of the value the previous year. So, r = 0.55. The first term

a1 corresponds to the year 2009, so the year 2014

corresponds to a6.

sequence to find the a6.

Therefore, the value of the equipment in 2014 is about $2516.42.

29. Find the sixth term of a geometric sequence with a first term of 9 and a common ratio of 2.

30. If r = 4 and a8 = 100, what is the first term of the geometric sequence?

SOLUTION:

Substitute a8 = 100, r = 4, and n = 8 into the

formula for the nth term of a geometric sequence

to find the a1.

31. X GAMES Refer to the beginning of the lesson. The X Games netted approximately $40 million in revenue in 2002. If the X Games continue to generate 13% more revenue each year, how much revenue will the X Games generate in 2020?

SOLUTION: The X Games netted about $40,000,000 in 2002, so

a1 = $40,000,000. Because the games generate

13% or 0.13 more revenue each year, the amount of revenue generated on a given year will be 1.13 times the revenue from the previous year. So, r =

1.13. The first term a1 corresponds to the year

2002, so the year 2020 corresponds to a19.

Therefore, the X Games will generate about $360.97 million in 2020.

Find the indicated geometric means for each pair of nonconsecutive terms.

32. 4 and 256; 2 means

SOLUTION: The sequence will resemble 4, __?__ , __?__ , 256. Note that a1 = 4, n = 4, and a4 = 256. Find the

common ratio using nth term for a geometric sequence formula.

The common ratio is 4. Use r to find the geometric means. 4(4) = 16 16(4) = 64 Therefore, a sequence with two geometric means between 4 and 256 is 4, 16, 64, 256.

33. 256 and 81; 3 means

SOLUTION: The sequence will resemble 256, __?__ , __?__ ,

__?__ , 81. Note that a1 = 256, n = 5, and a5 = 81.

Find the common ratio using the nth term for a geometric sequence formula.

The common ratio is . Use r to find the

geometric means.

r =

Therefore, a sequence with three geometric means between 256 and 81 is 256, –192, 144, –108, 81 or 256, 192, 144, 108, 81.

34. and 7; 1 mean

SOLUTION:

The sequence will resemble , __?__ , 7. Note

that a1 = , n = 3, and a3 = 7. Find the common

ratio using the nth term for a geometric sequence formula.

geometric means.

Therefore, a sequence with one geometric mean

between and 7 is , –2, 7 or , 2, 7.

35. –2 and 54; 2 means

SOLUTION: The sequence will resemble –2, __?__ , __?__ , 54. Note that a1 = –2, n = 4, and a4 = 54. Find the

The common ratio is –3. Use r to find the geometric means. –2(–3) = 6 6(–3) = –18 Therefore, a sequence with two geometric means between –2 and 54 is –2, 6, –18, 54.

36. 1 and 27; 2 means

37. 48 and −750; 2 means

SOLUTION: The sequence will resemble 48, __?__ , __?__ , –750. Note that a1 = 48, n = 4, and a4 = 48. Find the

geometric means.

= –120

= 300

Therefore, a sequence with two geometric means between 48 and –750 is 48, –120, 300, –750.

38. i and −1; 4 means

SOLUTION: The sequence will resemble i, __?__ , __?__ , __?__ , __?__ , –1. Note that a1 = i, n = 6, and a6 = –1. Find the

The common ratio is i. Use r to find the geometric means. i(i) = –1 –1(i) = –i –i(i) = 1 1(i) = i Therefore, a sequence with two geometric means

between i and −1 is i, –1, –i, 1, i, –1.

39. t 8 and t – 7; 4 means

SOLUTION:

The sequence will resemble t8, __?__ , __?__ ,

__?__ , __?__ , t – 7

.

Note that a1 = t8, n = 6, and a6 = t

– 7. Find the

geometric means.

= t5

= t2

= t–1

= t–4

Therefore, a sequence with two geometric means

between t8 and t

– 7 is t

8, t

5, t

2, t–1

, t–4

, t – 7

.

Find the sum of each geometric series described.

40. first six terms of 3 + 9 + 27 + …

SOLUTION: First, find the common ratio. 9 ÷ 3 = 3 27 ÷ 9 = 3 The common ratio is 3. Use Formula 1 for the sum of a finite geometric series.

41. first nine terms of 0.5 + (–1) + 2 + …

SOLUTION: First, find the common ratio. –1 ÷ 0.5 or –2 2 ÷ –1 or –2 The common ratio is –2. Use Formula 1 for the sum of a finite geometric series.

42. first eight terms of 2 + 2 + 6 + …

2 ÷ 2 or

6 ÷ 2 or

The common ratio is . Use Formula 1 for the sum of a finite geometric series.

43. first n terms of a1 = 4, an = 2000, r = −3

SOLUTION: Use Formula 2 for the nth partial sum of a geometric series.

44. [Contains update not in print edition] first n terms of a1 = 5, an = 1,310,720, r = 4

SOLUTION: [Solution for updated problem] Use Formula 2 for the nth partial sum of a geometric series.

45. first n terms of a1 = 3, an = 46,875, r = −5

46. first n terms of a1 = −8, an = −256, r = 2

47. first n terms of a1 = −36, an = 972, r = 7

Find each sum.

48.

SOLUTION:

Find n, a1, and r.

Substitute n = 6, a1 = 5, and r = 2 into the formula

for the sum of a finite geometric series.

49.

SOLUTION:

Find n, a1, and r.

Substitute n = 5, a1 = −4, and r = 3 into the formula

50.

SOLUTION:

Find n, a1, and r.

Substitute n = 5, a1 = 1, and r = −3 into the formula

51.

SOLUTION:

Find n, a1, and r.

Substitute n = 6, a1 = 2, and r = 1.4 into the

formula for the sum of a finite geometric series.

52.

SOLUTION:

Find n, a1, and r.

Substitute n = 6, a1 = 100, and r = into the

53.

SOLUTION:

Find n, a1, and r.

Substitute n = 9, a1 = , and r = −3 into the

54.

SOLUTION:

Find n, a1, and r.

55.

SOLUTION:

Find n, a1, and r.

If possible, find the sum of each infinite geometric series.

56. + + + …

÷ =

The common ratio r is 1. Therefore, this infinite geometric series has no sum.

58. 18 + (–27) + 40.5 + ...

SOLUTION: First, find the common ratio. –27 ÷ 18 = –1.5 40.5 ÷ –27 = –1.5 The common ratio r is > 1. Therefore, this infinite geometric series has no sum.

59. 12 + (–7.2) + 4.32 + ...

SOLUTION: First, find the common ratio. –7.2 ÷ 12 = –0.6 4.32 ÷ –7.2 = –0.6 The common ratio r is 1. Therefore, this infinite geometric series has no sum.

59. 12 + (–7.2) + 4.32 + ...

SOLUTION: First, find the common ratio. –7.2 ÷ 12 = –0.6 4.32 ÷ –7.2 = –0.6 The common ratio r is

1.

= –2

2.

and – .

3. 0.5, 0.75, 1.125, …

4. 8, 20, 50, …

5. 2x, 10x, 50x, …

6. 64x, 16x, 4x, …

16x ÷ 64x =

4x ÷ 16x =

x.

7. x + 5, 3x + 15, 9x + 45, …

8. –9 – y , 27 + 3y , –81 – 9y , …

d3, d4, d5.

πd5.

of the circles is

10. 36, 12, 4, …

12 ÷ 36 =

4 ÷ 12 =

a1 = 36, an =

11. 64, 16, 4, …

16 ÷ 64 =

4 ÷ 16 =

a1 = 64, an =

12. −2, 10, −50, …

a1 = –2, an =

13. 4, −12, 36, …

a1 = 4, an =

14. 4, 8, 16, …

a1 = 4, an =

15. 20, 30, 45, …

a1 = 20, an =

16. 15, 5, , …

5 ÷ 15 =

÷ 5 =

a1 = 15, an =

17. , , , …

÷ = 2

a1 = , an =

common ratio.

found in part a.

20. a9 for 60, 30, 15, …

30 ÷ 60 =

15 ÷ 30 =

21. a4 for 7, 14, 28, …

22. a5 for 3, 1, , …

1 ÷ 3 =

÷ 1 =

23. a6 for 540, 90, 15, …

90 ÷ 540 =

15 ÷ 90 =

24. a7 if a3 = 24 and r = 0.5

SOLUTION:

25. a6 if a3 = 32 and r = –0.5

SOLUTION:

26. a6 if a1 = 16,807 and r =

27. a8 if a1 = 4096 and r =

SOLUTION:

corresponds to a6.

SOLUTION:

to find the a1.

32. 4 and 256; 2 means

33. 256 and 81; 3 means

__?__ , 81. Note that a1 = 256, n = 5, and a5 = 81.

geometric means.

r =

34. and 7; 1 mean

SOLUTION:

geometric means.

35. –2 and 54; 2 means

36. 1 and 27; 2 means

37. 48 and −750; 2 means

geometric means.

= –120

= 300

39. t 8 and t – 7; 4 means

SOLUTION:

__?__ , __?__ , t – 7

.

– 7. Find the

geometric means.

= t5

= t2

= t–1

= t–4

between t8 and t

– 7 is t

8, t

5, t

2, t–1

, t–4

, t – 7

.

2 ÷ 2 or

6 ÷ 2 or

Find each sum.

48.

SOLUTION:

Find n, a1, and r.

49.

SOLUTION:

Find n, a1, and r.

50.

SOLUTION:

Find n, a1, and r.

51.

SOLUTION:

Find n, a1, and r.

52.

SOLUTION:

Find n, a1, and r.

53.

SOLUTION:

Find n, a1, and r.

54.

SOLUTION:

Find n, a1, and r.

55.

SOLUTION:

Find n, a1, and r.

56. + + + …

÷ =

58. 18 + (–27) + 40.5 + ...

59. 12 + (–7.2) + 4.32 + ...

1.

= –2

2.

and – .

3. 0.5, 0.75, 1.125, …

4. 8, 20, 50, …

5. 2x, 10x, 50x, …

6. 64x, 16x, 4x, …

16x ÷ 64x =

4x ÷ 16x =

x.

7. x + 5, 3x + 15, 9x + 45, …

8. –9 – y , 27 + 3y , –81 – 9y , …

d3, d4, d5.

πd5.

of the circles is

10. 36, 12, 4, …

12 ÷ 36 =

4 ÷ 12 =

a1 = 36, an =

11. 64, 16, 4, …

16 ÷ 64 =

4 ÷ 16 =

a1 = 64, an =

12. −2, 10, −50, …

a1 = –2, an =

13. 4, −12, 36, …

a1 = 4, an =

14. 4, 8, 16, …

a1 = 4, an =

15. 20, 30, 45, …

a1 = 20, an =

16. 15, 5, , …

5 ÷ 15 =

÷ 5 =

a1 = 15, an =

17. , , , …

÷ = 2

a1 = , an =

common ratio.

found in part a.

20. a9 for 60, 30, 15, …

30 ÷ 60 =

15 ÷ 30 =

21. a4 for 7, 14, 28, …

22. a5 for 3, 1, , …

1 ÷ 3 =

÷ 1 =

23. a6 for 540, 90, 15, …

90 ÷ 540 =

15 ÷ 90 =

24. a7 if a3 = 24 and r = 0.5

SOLUTION:

25. a6 if a3 = 32 and r = –0.5

SOLUTION:

26. a6 if a1 = 16,807 and r =

27. a8 if a1 = 4096 and r =

SOLUTION:

corresponds to a6.

SOLUTION:

to find the a1.

32. 4 and 256; 2 means

33. 256 and 81; 3 means

__?__ , 81. Note that a1 = 256, n = 5, and a5 = 81.

geometric means.

r =

34. and 7; 1 mean

SOLUTION:

geometric means.

35. –2 and 54; 2 means

36. 1 and 27; 2 means

37. 48 and −750; 2 means

geometric means.

= –120

= 300

39. t 8 and t – 7; 4 means

SOLUTION:

__?__ , __?__ , t – 7

.

– 7. Find the

geometric means.

= t5

= t2

= t–1

= t–4

between t8 and t

– 7 is t

8, t

5, t

2, t–1

, t–4

, t – 7

.

2 ÷ 2 or

6 ÷ 2 or

Find each sum.

48.

SOLUTION:

Find n, a1, and r.

49.

SOLUTION:

Find n, a1, and r.

50.

SOLUTION:

Find n, a1, and r.

51.

SOLUTION:

Find n, a1, and r.

52.

SOLUTION:

Find n, a1, and r.

53.

SOLUTION:

Find n, a1, and r.

54.

SOLUTION:

Find n, a1, and r.

55.

SOLUTION:

Find n, a1, and r.

56. + + + …

÷ =

58. 18 + (–27) + 40.5 + ...

59. 12 + (–7.2) + 4.32 + ...

1.

= –2

2.

and – .

3. 0.5, 0.75, 1.125, …

4. 8, 20, 50, …

5. 2x, 10x, 50x, …

6. 64x, 16x, 4x, …

16x ÷ 64x =

4x ÷ 16x =

x.

7. x + 5, 3x + 15, 9x + 45, …

8. –9 – y , 27 + 3y , –81 – 9y , …

d3, d4, d5.

πd5.

of the circles is

10. 36, 12, 4, …

12 ÷ 36 =

4 ÷ 12 =

a1 = 36, an =

11. 64, 16, 4, …

16 ÷ 64 =

4 ÷ 16 =

a1 = 64, an =

12. −2, 10, −50, …

a1 = –2, an =

13. 4, −12, 36, …

a1 = 4, an =

14. 4, 8, 16, …

a1 = 4, an =

15. 20, 30, 45, …

a1 = 20, an =

16. 15, 5, , …

5 ÷ 15 =

÷ 5 =

a1 = 15, an =

17. , , , …

÷ = 2

a1 = , an =

common ratio.

found in part a.

20. a9 for 60, 30, 15, …

30 ÷ 60 =

15 ÷ 30 =

21. a4 for 7, 14, 28, …

22. a5 for 3, 1, , …

1 ÷ 3 =

÷ 1 =

23. a6 for 540, 90, 15, …

90 ÷ 540 =

15 ÷ 90 =

24. a7 if a3 = 24 and r = 0.5

SOLUTION:

25. a6 if a3 = 32 and r = –0.5

SOLUTION:

26. a6 if a1 = 16,807 and r =

27. a8 if a1 = 4096 and r =

SOLUTION:

corresponds to a6.

SOLUTION:

to find the a1.

32. 4 and 256; 2 means

33. 256 and 81; 3 means

__?__ , 81. Note that a1 = 256, n = 5, and a5 = 81.

geometric means.

r =

34. and 7; 1 mean

SOLUTION:

geometric means.

35. –2 and 54; 2 means

36. 1 and 27; 2 means

37. 48 and −750; 2 means

geometric means.

= –120

= 300

39. t 8 and t – 7; 4 means

SOLUTION:

__?__ , __?__ , t – 7

.

– 7. Find the

geometric means.

= t5

= t2

= t–1

= t–4

between t8 and t

– 7 is t

8, t

5, t

2, t–1

, t–4

, t – 7

.

2 ÷ 2 or

6 ÷ 2 or

Find each sum.

48.

SOLUTION:

Find n, a1, and r.

49.

SOLUTION:

Find n, a1, and r.

50.

SOLUTION:

Find n, a1, and r.

51.

SOLUTION:

Find n, a1, and r.

52.

SOLUTION:

Find n, a1, and r.

53.

SOLUTION:

Find n, a1, and r.

54.

SOLUTION:

Find n, a1, and r.

55.

SOLUTION:

Find n, a1, and r.

56. + + + …

÷ =

58. 18 + (–27) + 40.5 + ...

59. 12 + (–7.2) + 4.32 + ...

1.

= –2

2.

and – .

3. 0.5, 0.75, 1.125, …

4. 8, 20, 50, …

5. 2x, 10x, 50x, …

6. 64x, 16x, 4x, …

16x ÷ 64x =

4x ÷ 16x =

x.

7. x + 5, 3x + 15, 9x + 45, …

8. –9 – y , 27 + 3y , –81 – 9y , …

d3, d4, d5.

πd5.

of the circles is

10. 36, 12, 4, …

12 ÷ 36 =

4 ÷ 12 =

a1 = 36, an =

11. 64, 16, 4, …

16 ÷ 64 =

4 ÷ 16 =

a1 = 64, an =

12. −2, 10, −50, …

a1 = –2, an =

13. 4, −12, 36, …

a1 = 4, an =

14. 4, 8, 16, …

a1 = 4, an =

15. 20, 30, 45, …

a1 = 20, an =

16. 15, 5, , …

5 ÷ 15 =

÷ 5 =

a1 = 15, an =

17. , , , …

÷ = 2

a1 = , an =

common ratio.

found in part a.

20. a9 for 60, 30, 15, …

30 ÷ 60 =

15 ÷ 30 =

21. a4 for 7, 14, 28, …

22. a5 for 3, 1, , …

1 ÷ 3 =

÷ 1 =

23. a6 for 540, 90, 15, …

90 ÷ 540 =

15 ÷ 90 =

24. a7 if a3 = 24 and r = 0.5

SOLUTION:

25. a6 if a3 = 32 and r = –0.5

SOLUTION:

26. a6 if a1 = 16,807 and r =

27. a8 if a1 = 4096 and r =

28. ACCOUNTING Julian Rockman is an accountantfor a small company. On January 1, 2009, the company purchased $50,000 worth of computers, printers, scanners, and hardware. Because this equipment

ms. wilson's math classes -...

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