Mathematics Advanced • Year 12 • Module 7 • Lesson 5
Geometric Sequences in Finance
Practise HSC-style writing on the GP-to-finance mapping — including a structured proof and projection.
1. Short-answer questions
1.1 $15,000 is invested at 4% p.a. compound annual. The year-end balances form a geometric sequence.
(a) State the first term a and common ratio r.
(b) Find the balance at the end of year 6 using Tₙ = arn−1. 3 marks Band 3
1.2 $25,000 worth of equipment depreciates at 10% p.a. reducing balance. The book values form a geometric sequence.
(a) State a and r.
(b) Find the book value at the end of year 4. 3 marks Band 3-4
1.3 $5,000 is invested at 8% p.a. compound annual for 8 years.
(a) Find S₈, the sum of the year-end balances from year 1 to year 8.
(b) Find the total interest earned over the 8 years.
(c) State, in one sentence, why annuity formulas in later lessons are derived from Sₙ and not from Tₙ. 4 marks Band 4
2. Extended response
2.1 A 30-year-old has $20,000 in a separate investment account. The account earns 6% per annum compounded annually. She makes no further deposits.
(a) Show that the year-end balances form a geometric sequence and state a and r.
(b) Prove, using the GP formula Tₙ = arn−1 with these specific a and r, that the year-n balance equals 20,000 × (1.06)ⁿ. Show the algebraic step that simplifies the exponent.
(c) Find the balance at age 60 and at age 70 to the nearest dollar.
(d) Use logarithms to find the age at which the balance first exceeds $200,000.
(e) Compute the sum S₃₀ of the first 30 year-end balances, and explain in one sentence what this sum represents financially (not the final balance). 10 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — states ratio test: each balance ÷ previous = 1.06, so the sequence is geometric.
• 1 mark — a = 20,000 × 1.06 = $21,200 (year-1 balance); r = 1.06.
Part (b) — 2 marks
• 1 mark — substitutes a and r into Tₙ = arn−1: Tₙ = 20,000(1.06) × (1.06)n−1.
• 1 mark — combines exponents: (1.06)1 + (n−1) = (1.06)ⁿ, hence Tₙ = 20,000(1.06)ⁿ.
Part (c) — 2 marks
• 1 mark — age 60 (n = 30): 20,000(1.06)³⁰ ≈ $114,870.
• 1 mark — age 70 (n = 40): 20,000(1.06)⁴⁰ ≈ $205,714.
Part (d) — 2 marks
• 1 mark — writes 20,000(1.06)ⁿ > 200,000 and isolates (1.06)ⁿ > 10.
• 1 mark — solves n > ln 10 / ln 1.06 ≈ 39.52 ⇒ first whole year is 40, hence age 70.
Part (e) — 2 marks
• 1 mark — correct S₃₀ ≈ $1,675,772.
• 1 mark — explicitly notes S₃₀ is the sum of all 30 year-end balances, not the final balance; useful e.g. for computing average annual carried value or for laying groundwork for annuity-style derivations.
Your response:
Stuck on (b)? Combine the exponents: 1 + (n − 1) = n.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $15,000 at 4% (3 marks)
Sample response. (a) a = 15,000 × 1.04 = $15,600; r = 1.04. (b) T₆ = 15,600 × (1.04)⁵ = 15,600 × 1.216653 = $18,979.79 (equivalently 15,000 × 1.04⁶).
Marking notes. 1 mark for (a). 2 marks for (b) — 1 mark for the formula substitution, 1 mark for the cent-level value. Acceptable to compute as 15,000 × 1.04⁶ directly with the equivalence noted.
1.2 — $25,000 at 10% RB (3 marks)
Sample response. (a) a = 25,000 × 0.90 = $22,500; r = 0.90. (b) T₄ = 22,500 × (0.90)³ = 22,500 × 0.729 = $16,402.50 (≡ 25,000 × 0.90⁴).
Marking notes. 1 mark for (a) — both values stated correctly. 2 marks for (b) — 1 for formula, 1 for the value. Common error: using r = 0.10 (the depreciation rate) instead of r = 0.90 (the common ratio).
1.3 — $5,000 at 8% for 8 years (4 marks)
(a) Sample. a = 5,000 × 1.08 = 5,400; r = 1.08. S₈ = 5,400(1.08⁸ − 1)/(1.08 − 1) = 5,400 × 0.85093 / 0.08 = $57,437.99.
(b) Sample. Total interest = S₈ − 8P = 57,437.99 − 40,000 = $17,437.99.
(c) Sample. An annuity makes a regular deposit each period, with each deposit then growing under compound interest; the future value is a sum of many compound-interest balances — exactly what Sₙ computes.
Marking notes. 2 marks for (a). 1 mark for (b). 1 mark for (c) — must mention "sum of many balances" or "each deposit grows".
2.1 — Extended response (10 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Ratio test and identification. Year-1 balance: 20,000 × 1.06 = $21,200. Year-2: 21,200 × 1.06 = $22,472. Year-3: 22,472 × 1.06 = $23,820.32. Ratio T₂/T₁ = 22,472/21,200 = 1.06; ratio T₃/T₂ = 23,820.32/22,472 = 1.06. The ratio is constant, so the year-end balances form a geometric sequence with a = $21,200, r = 1.06. [2 marks]
(b) Equivalence with the standard compound-interest formula. Substitute a = 20,000(1.06) and r = 1.06 into Tₙ = arn−1:
Tₙ = 20,000(1.06) × (1.06)n−1 = 20,000 × (1.06)1 + (n−1) = 20,000 × (1.06)ⁿ.
Hence Tₙ = 20,000(1.06)ⁿ, identical to the standard compound-interest formula. [2 marks — substitution + exponent-combining step.]
(c) Balances at ages 60 and 70. Age 60 (n = 30): T₃₀ = 20,000(1.06)³⁰ = 20,000 × 5.74349 ≈ $114,870. Age 70 (n = 40): T₄₀ = 20,000(1.06)⁴⁰ = 20,000 × 10.28572 ≈ $205,714. [2 marks]
(d) Age when balance first exceeds $200,000. Solve 20,000(1.06)ⁿ > 200,000 ⇒ (1.06)ⁿ > 10 ⇒ n > ln 10 / ln 1.06 = 2.3026 / 0.05827 ≈ 39.52. The first whole year is n = 40, hence age 30 + 40 = 70. [2 marks]
(e) Sum of first 30 year-end balances. S₃₀ = 21,200 × (1.06³⁰ − 1) / (1.06 − 1) = 21,200 × 4.74349 / 0.06 = 21,200 × 79.0582 = $1,675,734 (≈ $1.68 million). This is the sum of all 30 year-end balances — not the final balance (which is only $114,870). It is the value that underpins later annuity-style derivations, where each annual contribution sits in the account for a different number of years and the future value is a sum of those compounded amounts. [2 marks — calculation + interpretation.]
Total: 10/10.
Band descriptors for marker.
Band 3: (a) attempted with a and r stated but no ratio test; (b) restated rather than derived; (c) computed at one age; (d) attempted without logs; (e) missing or wrongly equated to T₃₀. ≈ 3-4 marks.
Band 4: (a)-(c) correct; (d) uses logarithms but doesn't interpret as "age 70"; (e) calculation correct but no interpretation. ≈ 6-7 marks.
Band 5: All parts numerically correct, exponent-combining step in (b) shown, interpretation in (e) attempts to distinguish sum from final balance. ≈ 8-9 marks.
Band 6: Full working at each step, explicit ratio test in (a), exponent-combining algebra in (b), age interpretation in (d), and (e) explicitly distinguishes Sₙ from Tₙ and previews the annuity application. 10/10.