Mathematics Advanced • Year 12 • Module 7 • Lesson 5

Geometric Sequences in Finance

Build fluency identifying GPs in financial contexts and applying Tₙ and Sₙ to compound interest and reducing-balance depreciation.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the GP formulas:

nth term: Tₙ = ____________________

sum of n terms (r ≠ 1): Sₙ = ____________________

Q1.2 For a compound-interest investment with principal P and rate i per period, write the common ratio r of the GP of year-end balances. (Hint: it is not i.)

Q1.3 State, in one sentence, the difference between Tₙ and Sₙ in finance.

Stuck? Revisit lesson § Formula Reference and § Misconceptions.

2. Worked example — $8,000 at 6% p.a., compound annually

Follow each line. Each step has a short reason.

Problem. $8,000 is invested at 6% p.a. compounded annually. (a) Find the year-7 balance T₇. (b) Find the sum S₇ of all year-end balances from year 1 to year 7.

Step 1 — Map the finance to a GP.

a = P(1 + i) = 8,000 × 1.06 = 8,480 (year-1 balance)

r = 1 + i = 1.06 (common ratio)

Reason: each year multiplies the previous balance by (1 + i), not by i alone.

Step 2 — Apply Tₙ = arⁿ⁻¹ with n = 7.

T₇ = 8,480 × (1.06)⁶ = 8,480 × 1.41852

T₇ ≈ $12,028.93

Step 3 — Apply Sₙ = a(rⁿ − 1)/(r − 1) with n = 7.

S₇ = 8,480 × (1.06⁷ − 1) / (1.06 − 1)

S₇ = 8,480 × (0.50363) / 0.06

S₇ ≈ $71,180.51

Step 4 — Interpret the answers.

T₇ is the balance at the end of year 7. S₇ is the total of every year-end balance from year 1 to year 7 — not the final balance.

3. Faded example — reducing-balance depreciation as a GP

An asset of V₀ = $50,000 depreciates at 12% p.a. reducing balance. Fill in each blank. 4 marks

Step 1 — Map to GP variables:

a = V₀(1 − d) = 50,000 × (1 − ______) = $______ (year-1 book value)

r = 1 − d = ______ (common ratio)

Step 2 — Write the first three terms (year-end book values):

T₁ = $______; T₂ = T₁ × ______ = $______; T₃ = T₂ × ______ = $______

Step 3 — Verify with the formula Tₙ = arⁿ⁻¹:

T₃ = ______ × (______)² = $______ (should match Step 2)

Conclusion. The book values form a GP with a = $______ and r = ______; year-3 book value is $______.

Stuck? Revisit lesson § Try It Now.

4. Graduated practice — Tₙ and Sₙ in finance

Foundation — identify and substitute (4 questions)

Q	Scenario	Working & answer
4.1 1	P = $1,000, i = 5% p.a. compound annual. State a and r of the GP.
4.2 1	Same GP as 4.1. Compute T₄ (year-4 balance).
4.3 1	V₀ = $20,000, d = 10% RB. State a and r of the GP of book values.
4.4 1	Same GP as 4.3. Compute T₃ (year-3 book value).

Standard — typical HSC difficulty (6 questions)

4.5 $15,000 is invested at 4% p.a. compounded annually. Find the GP common ratio and the year-6 balance using Tₙ = arⁿ⁻¹. 2 marks

4.6 $25,000 is invested at 10% p.a. compounded annually. Find the year-4 balance using the GP formula. 2 marks

4.7 $12,000 is invested at 6% p.a. compounded annually for 5 years. Find the sum S₅ of the year-end balances and the total interest earned. (Hint: total interest = S₅ − 5P.) 2 marks

4.8 An asset worth $40,000 depreciates at 15% p.a. RB. State a and r of the GP of book values, and use Tₙ to find the year-3 book value. 2 marks

4.9 $8,000 is invested at 10% p.a. compound annual. Use logs (or trial) to find the first year n for which Tₙ > $15,000. 2 marks

4.10 $5,000 is invested at 8% p.a. compounded annually for 8 years. Find S₈ (the sum of year-end balances) and the total interest. 2 marks

Extension — combine concepts (2 questions)

4.11 Sequence: 10,000; 10,500; 11,025; 11,576.25; … Show that this is geometric and find r. State which financial scenario produces this sequence. 3 marks

4.12 Show that for a compound-interest investment with principal P and rate i, the GP-formula Tₙ = arⁿ⁻¹ (with a = P(1+i), r = 1+i) is equivalent to the standard compound-interest formula A = P(1+i)ⁿ. 3 marks

Stuck on 4.12? Substitute a and r into Tₙ = arⁿ⁻¹ and simplify the exponents.

5. Self-check the easy 3

Tick once you have checked your method.

For 4.1 I wrote r = 1.05, not r = 0.05 — the common ratio is (1 + i).

For 4.4 I used r = 0.90 (RB) and obtained T₃ = 20,000 × 0.90² × 0.90 — the first term is a = V₀(1 − d), not V₀.

For 4.7 I used Sₙ (sum) — not Tₙ — because the question asks for the total of all balances.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — GP formulas

Tₙ = arⁿ⁻¹. Sₙ = a(rⁿ − 1) / (r − 1).

Q1.2 — Common ratio in finance

r = 1 + i (one plus the interest rate per period), not i itself — because each balance is the previous balance kept entirely plus the interest earned.

Q1.3 — Tₙ vs Sₙ

Tₙ is one balance (the nth term — e.g., the year-n balance). Sₙ is the sum of every term from T₁ to Tₙ — useful for annuity formulas, not for the final balance.

Q3 — Faded example: $50,000 at 12% RB

a = 50,000 × (1 − 0.12) = 50,000 × 0.88 = $44,000. r = 0.88. T₁ = $44,000; T₂ = 44,000 × 0.88 = $38,720; T₃ = 38,720 × 0.88 = $34,073.60. Verify: T₃ = 44,000 × (0.88)² = 44,000 × 0.7744 = $34,073.60. ✓

Q4.1 — P = $1,000, i = 5%

a = 1,000 × 1.05 = $1,050. r = 1.05.

Q4.2 — T₄ for the same GP

T₄ = 1,050 × (1.05)³ = 1,050 × 1.157625 = $1,215.51. (Equivalent to 1,000(1.05)⁴.)

Q4.3 — V₀ = $20,000, d = 10% RB

a = 20,000 × 0.90 = $18,000. r = 0.90.

Q4.4 — T₃ for the same GP

T₃ = 18,000 × (0.90)² = 18,000 × 0.81 = $14,580.

Q4.5 — $15,000 at 4% for 6 yr

r = 1.04. a = 15,000 × 1.04 = 15,600. T₆ = 15,600 × (1.04)⁵ = 15,600 × 1.216653 = $18,979.79. (Equivalent to 15,000(1.04)⁶.)

Q4.6 — $25,000 at 10% for 4 yr

a = 27,500. T₄ = 27,500 × (1.10)³ = 27,500 × 1.331 = $36,602.50. (= 25,000 × 1.10⁴.)

Q4.7 — $12,000 at 6% for 5 yr, sum S₅

a = 12,000 × 1.06 = 12,720; r = 1.06. S₅ = 12,720(1.06⁵ − 1)/0.06 = 12,720 × 0.33823 / 0.06 = 12,720 × 5.63709 = $71,711.75. Total interest = S₅ − 5P = 71,711.75 − 60,000 = $11,711.75.

Q4.8 — $40,000 at 15% RB

a = 40,000 × 0.85 = $34,000; r = 0.85. T₃ = 34,000 × (0.85)² = 34,000 × 0.7225 = $24,565.

Q4.9 — First n with Tₙ > $15,000 for $8,000 at 10%

Solve 8,000(1.10)ⁿ > 15,000 ⇒ (1.10)ⁿ > 1.875 ⇒ n > ln 1.875 / ln 1.10 = 0.6286 / 0.0953 = 6.59. First whole year: n = 7.

Q4.10 — $5,000 at 8% for 8 yr, sum S₈

a = 5,000 × 1.08 = 5,400; r = 1.08. S₈ = 5,400(1.08⁸ − 1)/0.08 = 5,400 × 0.85093 / 0.08 = $57,437.99. Total interest = 57,437.99 − 8 × 5,000 = $17,437.99.

Q4.11 — Identify the GP

Ratios: 10,500/10,000 = 1.05; 11,025/10,500 = 1.05; 11,576.25/11,025 = 1.05 — all equal, so the sequence is geometric with r = 1.05. Scenario: $10,000 invested at 5% p.a. compounded annually (with the initial $10,000 listed as T₀ rather than T₁).

Q4.12 — Equivalence of Tₙ and A = P(1 + i)ⁿ

Substitute a = P(1 + i) and r = 1 + i into Tₙ = arⁿ⁻¹:
Tₙ = P(1 + i) × (1 + i)ⁿ⁻¹ = P × (1 + i)^{1 + (n−1)} = P × (1 + i)ⁿ = A.
Hence the GP-formula and the standard compound-interest formula give the same year-n balance — they are the same identity, written in two notations.