Success TutoringWorksheets

Geometric Series in Action: Finance, Growth and Limiting Sums — Answers

Answer sheet with worked solutions for the Year 12 sequences & series worksheet.

Download answers PDFPrint answersBack to worksheet
Success TutoringWorksheetsAnswer sheetYear 12 · Mathematics · Sequences & Series

Geometric Series in Action: Finance, Growth and Limiting Sums — Answers

Part A: Terms and sums

A geometric sequence multiplies by the same common ratio r each time: a, ar, ar2, … The nth term is Tₙ = ar[sup n − 1].

The sum of the first n terms is Sₙ = [frac a(r[sup n] − 1)/r − 1] (convenient when r > 1) or Sₙ = [frac a(1 − r[sup n])/1 − r] (convenient when r < 1).

Worked example. 3 + 6 + 12 + … to 8 terms: a = 3, r = 2, S₈ = = 3 × 255 = 765.

1.Find the 7th term of the geometric sequence 5, 15, 45, …

Answer: 3645

  1. a = 5, r = 3, so T₇ = 5 × 36 = 5 × 729 = 3645
2.Find the sum of the first 10 terms of 2 + 4 + 8 + …

Answer: 2046

  1. a = 2, r = 2: S₁₀ = = 2 × 1023 = 2046
3.Find the sum of the first 6 terms of 64 + 48 + 36 + …

Answer: 210.4375

  1. a = 64, r = = 0.75
  2. S₆ = = 256 × (1 − 0.177978…)
  3. = 210.4375
4.How many terms of 3 + 6 + 12 + … are needed for the sum to exceed 10 000?

Answer: 12

  1. Sₙ = 3(2n − 1) > 10 000, so 2n > 3334.3…
  2. 211 = 2048 is too small and 212 = 4096 is enough.
  3. Check: S₁₁ = 6141 and S₁₂ = 12 285. 12 terms are needed.

Part B: Limiting sums

When −1 < r < 1, the terms shrink towards 0 and rn → 0, so the sum approaches the limiting sum S∞ = [frac a/1 − r]. If |r| ≥ 1 there is no limiting sum.

A recurring decimal is a geometric series: 0.333… = 0.3 + 0.03 + 0.003 + … with a = 0.3 and r = 0.1.

Worked example. 0.777… = = = [frac 7/9].

5.The limiting sum of 12 + 4 + + … is (answer space).

Answer: 18

a = 12 and r = , so S∞ = = 12 ÷ = 18.

6.The series 2 − 3 + 4.5 − … has a limiting sum.

Answer: False

The common ratio is r = −1.5, and |r| > 1, so the terms grow in size and the sum does not settle to a limit.

7.Write the recurring decimal 0.454545… as a fraction in simplest form, using a limiting sum.

Answer:

  1. 0.454545… = 0.45 + 0.0045 + 0.000045 + …, with a = 0.45 and r = 0.01
  2. S∞ = = =
  3. = [frac 5/11]
8.The series 1 + 2x + 4x2 + … has a limiting sum when − < x < (answer space), and when x = 0.2 its limiting sum is (answer space).

Answer: ,

The ratio is r = 2x, and a limiting sum needs |2x| < 1, that is − < x < . When x = 0.2, r = 0.4 and S∞ = = = .

Part C: The future value of regular deposits

If $M is deposited at the end of each period at interest rate r per period, then after n deposits the first deposit has grown to M(1 + r)n − 1 and the last has earned nothing. The balance is the geometric series M + M(1 + r) + … + M(1 + r)n − 1, so FV = [frac M((1 + r)[sup n] − 1)/r].

Worked example. $100 at the end of each month at 0.5% per month for 12 months: FV = = $1233.56. Only $1200 was deposited, so $33.56 is interest.

9.$2000 is deposited at the end of each year for 10 years into an account earning 5% p.a. compounded annually. Find the balance just after the 10th deposit.

Answer: $25 155.79

  1. The last deposit earns no interest, the first earns 9 years: FV = 2000 + 2000(1.05) + … + 2000(1.05)9
  2. This is a geometric series with a = 2000, r = 1.05, n = 10.
  3. FV = = 2000 × 12.5778… = $25 155.79
10.$250 is deposited at the end of each month for 5 years into an account earning 6% p.a. compounded monthly. Find the final balance.

Answer: $17 442.51

  1. r = 0.06 ÷ 12 = 0.005 per month; n = 60 deposits
  2. FV = = 250 × 69.7700…
  3. = $17 442.51
11.What equal deposit at the end of each month is needed to reach $50 000 in 8 years, at 4.8% p.a. compounded monthly?

Answer: $428.25

  1. r = 0.048 ÷ 12 = 0.004; n = 96
  2. 50 000 = , so M = =
  3. M = $428.25 (rounded to the cent)
12.Which expression equals $500 + $500(1.03) + $500(1.03)2 + … + $500(1.03)19?
  1. A
  2. B
  3. C500 × 1.0320
  4. D

Answer:

The powers run from 0 to 19, so there are 20 terms, with a = 500 and r = 1.03: S₂₀ = . There is no limiting sum because r > 1.

Part D: Loans and growth

For a reducing-balance loan of $P at rate r per period with repayment $M, the balance after n repayments is P(1 + r)n − M. Setting it to 0 gives M = [frac Pr(1 + r)[sup n]/(1 + r)[sup n] − 1].

Anything that grows (or shrinks) by a fixed percentage each step forms a geometric sequence, so totals are geometric series.

Worked example. A ball dropped from 2 m rises to half its height each bounce. Distance = 2 + 2(1 + 0.5 + 0.25 + …) = 2 + 2 × = 6 m.

13.$20 000 is borrowed at 6% p.a. compounded monthly and repaid in equal monthly repayments over 3 years. Using a geometric series, find the monthly repayment.

Answer: $608.44

  1. r = 0.005, n = 36. After 36 repayments of $M the balance is 20 000(1.005)36 − M(1 + 1.005 + … + 1.00535) = 0
  2. The bracket is a geometric series:
  3. M = =
  4. M = $608.44
14.A ball is dropped from a height of 10 m. After each bounce it rises to 60% of the height it fell from. Find the total vertical distance it travels before coming to rest.

Answer: 40 m

  1. Down 10 m, then up and down 6 m, 3.6 m, 2.16 m, …
  2. Distance = 10 + 2(6 + 3.6 + 2.16 + …) = 10 + 2 ×
  3. = 10 + 2 × 15 = 40 m
15.A job pays $60 000 in the first year, and the salary rises by 3% each year. Find the total earned over the first 10 years, to the nearest cent.

Answer: $687 832.76

  1. Salaries: 60 000, 60 000(1.03), …, 60 000(1.03)9
  2. S₁₀ = = 60 000 × 11.4638…
  3. = $687 832.76
16.For the same job, in which year does the salary first exceed $80 000?

Answer: 11

  1. Year n salary = 60 000 × 1.03n − 1 > 80 000, so 1.03n − 1 >
  2. n − 1 > = 9.73…, so n − 1 = 10
  3. The salary first exceeds $80 000 in year 11 (it is $80 634.98 that year).

Part E: Extended problem

Alex deposits $300 at the end of every month into a savings account that earns 4.2% p.a. compounded monthly (0.35% per month). Let Aₙ be the balance just after the nth deposit.

(a) Find A₂ and A₃, and show that Aₙ = .

(b) Find the balance after 4 years.

(c) How much of that balance is interest?

(d) After how many months will the balance first reach $25 000?

(e) What single amount invested now, at the same rate, would grow to the same balance as in part (b) after 4 years?

17.Answer parts (a) to (e), showing full working.[11 marks]

Model answer: (a) After 1 month: A₁ = 300. After 2 months: A₂ = 300(1.0035) + 300 = $601.05. After 3 months: A₃ = 601.05(1.0035) + 300 = $903.15 = 300(1 + 1.0035 + 1.00352). In general Aₙ = 300(1 + 1.0035 + … + 1.0035n − 1) = . (b) A₄₈ = = $15 650.54. (c) Deposits: 48 × 300 = $14 400, so interest = $1250.54. (d) ≥ 25 000 gives 1.0035n ≥ 1.291666…, so n ≥ = 73.25…, so 74 months (A₇₃ = $24 902.67 and A₇₄ = $25 289.83). (e) PV = = $13 234.13.

Success criteria:

  • (a) finds A₂ and A₃ and identifies the geometric series (3 marks)
  • (b) $15 650.54 (2 marks)
  • (c) $14 400 deposited and $1250.54 interest (1 mark)
  • (d) uses logarithms or checks values to get 74 months (3 marks)
  • (e) discounts the balance to get $13 234.13 (2 marks)
Free worksheets at worksheets.successtutoring.com · © Success Tutoring · Free for personal and classroom use

Marking tips

  • Give credit for correct method even when the final answer slips — the working shows where the thinking went right.
  • For open-ended tasks, use the success criteria as a checklist rather than looking for one "right" answer.
  • Celebrate what went well first, then pick one thing to work on next.

Great work. Here's what to try next: