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:
$10,500 after year 1, $11,025 after year 2, $11,576.25 after year 3, …
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 (from the initial $10,000 to year 1: 500; from year 1 to year 2: 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
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 S₅, the sum of the five year-end book values, and state what this sum represents (and why it is not the asset balance at any one date). 3 marks
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 on this single deposit (T₈ − P); S₈ is the sum of balances, not the interest or the account balance. 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
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) 11,025/10,500 = 1.05; 11,576.25/11,025 = 1.05; 12,155.0625/11,576.25 = 1.05, common ratio is constant, so the sequence is geometric with r = 1.05.
(ii) Differences: 525, 551.25, 578.8125, 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) Taking T₁ = $10,500 as the balance after year 1, T₁₂ = 10,500 × 1.05¹¹ = $17,958.56 at the end of year 12.
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 = $98,357.57. Age 55 (n = 20): T20 = 50,000(1.07)²⁰ = $193,484.22. Age 65 (n = 30): T30 = 50,000(1.07)³⁰ = $380,612.75.
(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. The lump sum remains below $500,000 at year 30 and first exceeds it at year 35.
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) = $126,093.46. This is the sum of five year-end book-value snapshots, not a book value at one date. Their arithmetic mean is S₅ ÷ 5 = $25,218.69.
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 = $57,437.79, the sum of the year-end balances.
(ii) T₈ = 5,000 × 1.08⁸ = $9,254.65, so interest on the single deposit is T₈ − P = 9,254.65 − 5,000 = $4,254.65.
(iii) For comparison, an ordinary annuity has equal deposits made at the end of each period. Each deposit compounds for the periods remaining after it is made. 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) After year 1 the balances are equal at $10,500. The exponential balance first exceeds the linear balance after year 2: L = $11,000 and E = 10,000 × 1.05² = $11,025.
(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.