Mathematics Advanced • Year 12 • Module 7 • Lesson 5
Geometric Sequences in Finance
Apply GP term and sum formulas to superannuation projections, depreciation timelines and the build-up to annuities.
Problem 1 — Is this sequence geometric? (the "Think First" scenario)
$10,000 is invested at 5% p.a. compound interest. The year-end balances form the sequence:
$10,000, $10,500, $11,025, $11,576.25, …
Set up: What are we solving for?
(i) Show, by the ratio test, that the sequence is geometric and state the common ratio r. 2 marks
(ii) Show that the differences (10,500 − 10,000 = 500, 11,025 − 10,500 = 525, …) are not constant — hence the sequence is not arithmetic. 2 marks
(iii) Use the GP formula Tₙ = arn−1 to find the balance at the end of year 12. 2 marks
Stuck? Revisit lesson § Revisit Your Initial Thinking.Problem 2 — Superannuation projection (the "Real-World Anchor")
A 35-year-old has $50,000 in super today. The fund earns approximately 7% p.a.
Set up: What are we solving for?
(i) State the GP first term a and common ratio r if Tn denotes the year-n balance (after compounding) of this lump sum, assuming no additional contributions. 1 mark
(ii) Use the GP formula to project the balance at age 45, 55 and 65 (10, 20 and 30 years from now). 3 marks
(iii) Use logarithms to find the age at which the balance first exceeds $500,000 (no additional contributions). 3 marks
Problem 3 — Depreciation as a GP (reducing balance)
An asset of V₀ = $40,000 depreciates at 15% per year on a reducing-balance basis.
Set up: What are we solving for?
(i) State the GP first term a and common ratio r for the year-end book values. 1 mark
(ii) Use Tₙ to find the year-5 book value. 2 marks
(iii) Calculate the sum of the first five year-end book values (S₅). Why is this sum financially relevant when accountants estimate the asset's "total carried value over five years"? 3 marks
Stuck? Use Sₙ = a(rⁿ − 1)/(r − 1) with r = 0.85.Problem 4 — Sum of balances and the build-up to annuities
$5,000 is invested at 8% p.a. compounded annually for 8 years.
Set up: What are we solving for?
(i) Find S₈, the sum of the year-end balances from year 1 to year 8. 2 marks
(ii) Find the total interest earned over the term using the relationship total interest = Sₙ − nP. 2 marks
(iii) Annuity (future-value) formulas, which you meet in lessons 7-10, are derived from Sₙ — not Tₙ. Explain in 2-3 sentences why a regular-deposit annuity is mathematically a sum of compound-interest balances rather than a single balance. 3 marks
Problem 5 — Linear vs exponential growth shape
An investor compares two scenarios on $10,000:
Scenario L: deposits an additional $500 in cash each year (linear addition, no interest).
Scenario E: no additional deposits, but the original $10,000 grows at 5% p.a. compound annual.
Set up: What are we solving for?
(i) Tabulate the balances at years 5, 10, 15, 20 for each scenario. (Scenario L: 10,000 + 500n. Scenario E: 10,000 × 1.05ⁿ.) 3 marks
(ii) Identify the first whole year in which Scenario E (exponential) overtakes Scenario L (linear). 2 marks
(iii) Explain in one sentence what the lesson's "is the curve linear, exponential or logarithmic?" Part 2 question is really asking. 2 marks
Stuck? Compare growth shapes: linear adds the same amount each year; exponential multiplies by a fixed factor.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Identify the GP
Set up. We are confirming the sequence is geometric by the ratio test, ruling out arithmetic, and projecting forward.
(i) 10,500/10,000 = 1.05; 11,025/10,500 = 1.05; 11,576.25/11,025 = 1.05 — common ratio is constant, so the sequence is geometric with r = 1.05.
(ii) Differences: 500, 525, 551.25 — not constant, so the sequence is not arithmetic. (For arithmetic the common difference must be constant; for geometric the common ratio must be constant.)
(iii) Treating T₁ = 10,000, T₂ = 10,500, … with a = 10,000, r = 1.05: T₁₂ = 10,000 × (1.05)¹¹ = 10,000 × 1.71034 = $17,103.39. (Equivalent to 10,000 × 1.05¹² if we label T₀ = 10,000 instead.)
Problem 2 — Superannuation projection
Set up. We are modelling super as a single lump sum compounding at 7% and locating both fixed-year balances and a balance-target year.
(i) a = 50,000 × 1.07 = $53,500 (year-1 balance); r = 1.07.
(ii) Age 45 (n = 10): T10 = 50,000(1.07)¹⁰ = 50,000 × 1.96715 = $98,357.57. Age 55 (n = 20): 50,000 × 3.86968 = $193,484.22. Age 65 (n = 30): 50,000 × 7.61226 = $380,613.02.
(iii) Solve 50,000(1.07)ⁿ > 500,000 ⇒ (1.07)ⁿ > 10 ⇒ n > ln 10 / ln 1.07 = 2.3026 / 0.06766 = 34.03 years. First year above $500,000 is year 35, i.e. age 70. (This is why super funds emphasise regular contributions — a lump sum at 7% cannot quite hit $500k in 30 years from $50k.)
Problem 3 — Depreciation GP
Set up. We are mapping reducing-balance depreciation to a GP and using both Tₙ and Sₙ.
(i) a = 40,000 × (1 − 0.15) = 40,000 × 0.85 = $34,000; r = 0.85.
(ii) T₅ = 34,000 × (0.85)⁴ = 34,000 × 0.52200625 = $17,748.21 (≡ 40,000 × 0.85⁵).
(iii) S₅ = 34,000 × (0.85⁵ − 1) / (0.85 − 1) = 34,000 × (−0.5563) / (−0.15) = 34,000 × 3.70905 = $126,107.71. Accountants use S₅ to estimate the area under the book-value curve — the total carried value of the asset across five years — which feeds into measures like average depreciation expense, internal cost recovery and (in more advanced applications) the present value of an asset's service over its life.
Problem 4 — Sum of balances and annuities
Set up. We are practising the Sₙ formula and then conceptually previewing how annuities are derived.
(i) a = 5,000 × 1.08 = 5,400; r = 1.08. S₈ = 5,400 × (1.08⁸ − 1)/0.08 = 5,400 × 0.85093 / 0.08 = 5,400 × 10.6366 = $57,437.99.
(ii) Total interest = S₈ − 8P = 57,437.99 − 8 × 5,000 = 57,437.99 − 40,000 = $17,437.99.
(iii) An annuity makes the same regular payment at the end of each period; each payment then earns compound interest for the remaining periods. The future value of the annuity is the sum of these many compound-interest contributions — payment 1 grows for the longest, payment n for the shortest. That sum is exactly a GP sum (with first term equal to the last payment and common ratio (1 + i)), which is why Sₙ — not Tₙ — appears in annuity derivations.
Problem 5 — Linear vs exponential growth
Set up. We are tabulating both scenarios to see the qualitative difference between arithmetic and geometric growth.
(i) Table values:
| n | Scenario L (linear) | Scenario E (exponential) |
|---|---|---|
| 5 | $12,500 | $12,762.82 |
| 10 | $15,000 | $16,288.95 |
| 15 | $17,500 | $20,789.28 |
| 20 | $20,000 | $26,532.98 |
(ii) Exponential overtakes by year 5 already (12,762.82 vs 12,500). The first whole year is n = 4: L = $12,000, E = 10,000 × 1.05⁴ = $12,155.06.
(iii) The lesson's "is the curve linear, exponential or logarithmic?" prompt is really checking whether students recognise that compound-interest balances are exponential (each year is multiplied by 1 + i), not linear (where each year adds the same dollar amount). That distinction is the whole point of GPs in finance.