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Module 7 · L5 of 20 ~40 min ⚡ +100 XP available

Geometric Sequences in Finance

Compound interest and depreciation are not new formulas — they are geometric sequences in disguise. Once you see the connection, you unlock the full power of GP sum formulas to calculate total interest, accumulated balances, and depreciation in a single step.

Today's hook — Your super fund projects a $50,000 balance growing to $380,000 over 30 years at 7% p.a. They didn't use magic — they used $T_{30}$ of a geometric sequence with $r = 1.07$. Every balance projection you'll ever see is a GP term.

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Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

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Think first — your gut answer

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You invest $10,000 at 5% p.a. compound interest. The balances at the end of each year form a sequence:

$10,000 → $10,500 → $11,025 → $11,576.25 → ...

Without using a formula — is this sequence arithmetic, geometric, or neither? How can you tell?

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The GP connection — one insight to rule them all

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Compound interest and reducing balance depreciation are geometric sequences in disguise. Lock this mapping and the rest of the module is just substitution.

When you invest $P at rate $i$ per period, each balance is the previous balance multiplied by $(1+i)$. That constant multiplier is the common ratio of a GP — which means every compound-interest formula is just $T_n = ar^{n-1}$ in disguise.

$$T_n = ar^{n-1} \quad \Longleftrightarrow \quad A = P(1+i)^n$$

Compound interest GP

$a = P(1+i)$ · $r = (1+i)$ · the rate is not $r$ — the ratio is $(1+i)$.

Depreciation GP

$a = V_0(1-d)$ · $r = (1-d)$ · values shrink because $r < 1$.

$T_n$ vs $S_n$

$T_n$ = one balance. $S_n$ = sum of all balances. Don't confuse them — exam trap!

What you'll master

Know

Key facts

Compound interest balances form a GP with $r = (1+i)$
Reducing balance depreciation forms a GP with $r = (1-d)$
The GP sum formula gives the total of all balances, not the final balance

Understand

Concepts

Why the common ratio in finance is $(1+\text{rate})$, not the rate itself
How $S_n$ relates to total interest earned across all periods
The difference between $T_n$ (one term) and $S_n$ (sum of terms)

Can do

Skills

Identify a financial scenario as a GP and state $a$ and $r$
Use $T_n$ to find any year's balance
Use $S_n$ to find total value accumulated over multiple periods

Key terms

Geometric sequence (GP)A sequence where each term is found by multiplying the previous term by a constant ratio $r$.

Common ratio $r$The constant multiplier between consecutive terms. For compound interest, $r = (1+i)$.

$T_n = ar^{n-1}$The $n$-th term formula. Gives one specific balance or value.

$S_n = \frac{a(r^n-1)}{r-1}$The sum of first $n$ terms. Gives the total of all balances accumulated.

Reducing balance depreciationA fixed percentage $d$ applied to the current value each period: $r = (1-d)$.

Principal $P$The initial amount invested or borrowed. Corresponds to $a / (1+i)$ in the GP mapping.

Compound interest as a geometric sequence

core concept

When you invest $P at rate $i$ per period, the balance at the end of each period is:

Period 1: $P(1+i)$ · Period 2: $P(1+i)^2$ · Period 3: $P(1+i)^3$ · ...

This is a geometric sequence with first term $a = P(1+i)$ and common ratio $r = (1+i)$.

$$T_n = ar^{n-1} = P(1+i) \cdot (1+i)^{n-1} = P(1+i)^n$$

Critical insight: The common ratio is not the interest rate $i$. It is $(1+i)$. Each balance is the previous balance plus the interest — you keep 100% and add the interest. If you used $r = 0.05$ for 5% interest, the sequence would shrink: $10{,}000,\ 500,\ 25,\ldots$ — completely wrong.

Superannuation projections. When your super fund projects your retirement balance, it is calculating $T_n$ of a GP. $50,000 today growing at 7% p.a. for 30 years is simply $T_{30}$ with $a = 50{,}000 \times 1.07$ and $r = 1.07$. That gives $\$50{,}000 \times 1.07^{30} \approx \$380{,}613$ — every superannuation projection is a GP term.

Compound interest GP: $a = P(1+i)$, $r = (1+i)$, so $T_n = P(1+i)^n$ (balance after $n$ periods); Common ratio is $(1+i)$ — never just $i$. Using $r = i$ will give a collapsing sequence.

Pause — copy the GP model for compound interest: first term $a = P(1+i)$, common ratio $r = (1+i)$, $n$th term $T_n = P(1+i)^n$ — noting that $r = (1+i)$, not just $i$ — into your book.

Quick check: An investment earns 6% p.a. compound interest. What is the common ratio $r$ of the GP formed by the year-end balances?

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BALANCE USING $T_n$

An investment of $8,000 earns 6% p.a. compounded annually. Find the balance at the end of year 7.

$a = 8{,}000(1.06) = 8{,}480$, $\quad r = 1.06$

Identify the GP. The first term $a$ is the balance after period 1.

PROBLEM 2 · SUM FORMULA $S_n$

Find the total of all year-end balances from year 1 to year 7 for the investment above ($8,000 at 6% p.a.).

$a = 8{,}480$, $r = 1.06$, $n = 7$

Same GP as before — now we sum all 7 terms.

PROBLEM 3 · DEPRECIATION GP

A car is valued at $50,000. It depreciates at 12% p.a. reducing balance. Write the first three book values and find the value after 5 years.

$a = 50{,}000(0.88) = 44{,}000$, $\quad r = 0.88$

$d = 12\%$, so $r = 1 - 0.12 = 0.88$. First book value after 1 year.

Did you get this? True or false: $S_n$ in the GP sum formula gives the final balance after $n$ periods of compound interest.

Common errors · the 3 traps that cost marks

Trap 01

Using $r = i$ instead of $r = (1+i)$

Using the interest rate directly as the common ratio gives a collapsing sequence. Always add 1: $r = 1 + i$ for growth, $r = 1 - d$ for depreciation.

Trap 02

Confusing $T_n$ with $S_n$

$T_n$ is the single balance at period $n$. $S_n$ is the sum of all balances up to period $n$. An exam question asking "find the balance" wants $T_n$, not $S_n$.

Trap 03

Wrong power in $T_n = ar^{n-1}$

For the balance at the end of year 7, the power is $r^6$ (not $r^7$). The formula is $T_n = ar^{n-1}$. Alternatively, use $A = P(1+i)^n$ directly to avoid this confusion.

Three-things-learned: Name three things that are true about the common ratio in a compound interest GP.

Odd one out: Three of these statements about GP finance are correct. Which one is wrong?

Quick-fire drill · 5 GP finance problems

$12,000 at 6% p.a. Write $a$ and $r$ for the compound interest GP.

Use $T_4 = ar^3$ to find the balance at end of year 4 (P = $12,000, i = 6%).

$8,000 at 10% p.a. At what year does the balance first exceed $15,000?

Car worth $40,000 depreciates at 15% p.a. Write $a$ and $r$.

Find the total of all year-end balances (years 1–5) for $10,000 at 4% p.a. Use $S_5$.

Fill in the blanks: For reducing balance depreciation at rate $d$, the common ratio is r = (1 __ d), which means $r$ is __ than 1, and the sequence is __easing (increasing / decreasing).

Revisit your thinking

The sequence $10{,}000,\ 10{,}500,\ 11{,}025,\ 11{,}576.25,\ldots$ is geometric. Test: $10{,}500/10{,}000 = 1.05$ and $11{,}025/10{,}500 = 1.05$ — constant ratio. It is not arithmetic because the differences are not constant ($500,\ 525,\ 551.25,\ldots$). The ratio test is the definitive check for geometric sequences.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next.

Short answer

ApplyBand 43 marks

Q1. An investment of $15,000 earns 4% p.a. compounded annually. (a) Write the first term $a$ and common ratio $r$ for the GP of year-end balances. (b) Find the balance at the end of year 6. Show working. (3 marks)

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ApplyBand 43 marks

Q2. A piece of machinery valued at $25,000 depreciates at 10% p.a. reducing balance. (a) Write $a$ and $r$. (b) Find the book value at the end of year 4. (3 marks)

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AnalyseBand 54 marks

Q3. $5,000 is invested at 8% p.a. for 8 years. (a) Use the GP sum formula to find the total of all year-end balances. (b) Calculate the total interest earned across all 8 years (i.e. total balances minus total principal). (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $a = 12{,}720$, $r = 1.06$ · 2: $T_4 = 12{,}720(1.06)^3 = \$15{,}149.72$ · 3: Year 7 (first exceeds $15,000) · 4: $a = 34{,}000$, $r = 0.85$ · 5: $a = 10{,}400$, $S_5 = 10{,}400(1.04^5-1)/0.04 = \$56{,}329$

Q1 (3 marks): (a) $a = 15{,}000(1.04) = 15{,}600$, $r = 1.04$ [1]. (b) $T_6 = 15{,}600(1.04)^5 = \$18{,}979.78$ [2].

Q2 (3 marks): (a) $a = 25{,}000(0.90) = 22{,}500$, $r = 0.90$ [1]. (b) $T_4 = 22{,}500(0.90)^3 = \$16{,}402.50$ [2].

Q3 (4 marks): (a) $a = 5{,}000(1.08) = 5{,}400$; $S_8 = 5{,}400(1.08^8 - 1)/0.08 = \$58{,}046.22$ [2]. (b) Total principal = $8 \times 5{,}000 = 40{,}000$. Total interest = $58{,}046.22 - 40{,}000 = \$18{,}046.22$ [2].

Boss battle · The Banker

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Five timed questions on GP finance. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering GP finance questions. Pool: lessons 1–5.

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← Lesson 4 · Depreciation Lesson 6 · Interest Rate Conversions →

Module overview · Maths Advanced · Checkpoint 1