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.
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
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 formulais this sequence arithmetic, geometric, or neither? How can you tell?
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.
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
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)
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
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 $r = (1+i)$.
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.
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
An investment of $8,000 earns 6% p.a. compounded annually. Find the balance at the end of year 7.
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 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.
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
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).
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.
Pick your answer, then rate your confidencethat tells the system what to drill next.
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)
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)
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)
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].
Five timed questions on GP finance. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Climb platforms by answering GP finance questions. Pool: lessons 1–5.
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