Compound interest loans and investments: QCE General Mathematics Unit 4 Loans, investments and annuities 1

Syllabus dot point

“Use the recurrence relation A(n+1) = r A(n), with r = 1 + i, and the compound interest formula A = P(1 + i)^n to model compound interest loans and investments, and solve practical problems for the total amount, total interest, principal, interest rate and number of compounding periods”

QCEGeneral MathematicsUnit 4: Investing and networking15 min readUpdated 23 Sep 2026

Quick answer

A compound interest loan or investment is modelled by $A_{n+1} = rA_n$ with $r = 1 + i$ , or by $A = P(1 + i)^n$ , where $i$ is the rate per compounding period and $n$ the number of periods. Interest is $A - P$ ; the principal is $\frac{A}{(1+i)^n}$ ; the rate comes from $1 + i = \left(\frac{A}{P}\right)^{1/n}$ ; and the number of periods is found by trial and improvement. Higher rates, longer times and more frequent compounding all increase the amount.

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The answer
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What this dot point is asking

Compound interest is interest calculated on the current balance, which includes interest added earlier. This sub-topic opens Unit 4 of the QCAA General Mathematics 2025 syllabus. You must model a compound interest investment or loan in two ways:

with the recurrence relation $A_{n+1} = rA_n$ , where $r = 1 + i$ and $i$ is the interest rate per compounding period, and
with the compound interest formula $A = P(1 + i)^n$ .

You then solve practical problems: finding the total amount, the total interest, the principal, the interest rate per year and per compounding period, the number of compounding periods, and explaining the effect of the interest rate and the number of compounding periods. (The effective annual rate, the other part of this sub-topic, has its own page.)

Remember the QCAA examination conditions: you have a scientific calculator and the QCAA formula book, not a CAS finance solver. Everything on this page is done with the formula, the recurrence and careful calculator use.

The answer

Rate per compounding period

Interest rates are quoted per annum (p.a.), but interest is added at the end of each compounding period, which may be a year, six months, a quarter, a month or a day. Convert before you do anything else:

i = \frac{\text{annual rate (as a decimal)}}{\text{number of compounding periods per year}}, \qquad n = \text{years} \times \text{periods per year}.

Compounding	Periods per year	4.8% p.a. gives $i =$	3 years gives $n =$
Annually	1	0.048	3
Six-monthly	2	0.024	6
Quarterly	4	0.012	12
Monthly	12	0.004	36
Daily	365	about 0.0001315	1095

The recurrence relation

Each period the amount is multiplied by $r = 1 + i$ :

A_{n+1} = rA_n, \qquad A_0 = P.

Here $A_n$ is the amount after $n$ compounding periods and $A_0$ is the principal. For $12 000 invested at 4.8% p.a. compounding monthly, $r = 1.004$ :

A_1 = 1.004 \times 12\,000 = 12\,048.00, \quad A_2 = 1.004 \times 12\,048.00 = 12\,096.19, \quad A_3 = 12\,144.58.

The recurrence shows the process step by step and is ideal for a table or a spreadsheet. It also shows why compound interest is geometric: each term is the previous term times the same ratio $r$ , so the amounts form a geometric sequence (see the sequences page in Unit 3).

The compound interest formula

Applying the multiplier $n$ times gives the formula directly:

A = P(1 + i)^n,

where $A$ is the total amount, $P$ is the principal, $i$ is the interest rate per compounding period and $n$ is the number of compounding periods. The total interest is

\text{interest} = A - P.

For the same investment after 5 years ( $n = 60$ ): $A = 12\,000(1.004)^{60} \approx 15\,247.69$ , so the investment earns $3247.69 in interest.

Key fact

Compound interest: $A_{n+1} = rA_n$ with $r = 1 + i$ , or $A = P(1 + i)^n$ . Always convert to $i$ (rate per compounding period) and $n$ (number of compounding periods) first. Interest $= A - P$ . Rearrangements: $P = \dfrac{A}{(1 + i)^n}$ and $1 + i = \left(\dfrac{A}{P}\right)^{1/n}$ . Find $n$ by trial and improvement or a table.

Loans and investments

The mathematics is identical: an investment grows as interest is added to it; a compound interest loan (with no repayments until the end) grows as interest is charged on it. A $5000 loan at 9.6% p.a. compounding monthly, repaid in one payment after 2 years, grows to $5000(1.008)^{24} \approx 6053.73$ , so the borrower pays $1053.73 in interest. Loans with regular repayments are reducing balance loans, which use a different recurrence ( $A_{n+1} = rA_n - d$ ) and are covered on the loans page.

Solving for other quantities

Principal: Rearrange to $P = \dfrac{A}{(1 + i)^n}$ . To have $20 000 in 6 years at 5.2% p.a. compounding quarterly ( $i = 0.013$ , $n = 24$ ): $P = \dfrac{20\,000}{1.013^{24}} \approx 14\,669.09$ .
Interest rate: Rearrange to $(1 + i)^n = \dfrac{A}{P}$ and take the $n$ th root: $1 + i = \left(\dfrac{A}{P}\right)^{1/n}$ . If $8000 grows to $9500 in 4 years compounding annually, $1 + i = (1.1875)^{1/4} \approx 1.0439$ , a rate of about 4.39% p.a. If the compounding is monthly, the root gives the monthly rate; multiply by 12 for the nominal annual rate.
Number of periods: The syllabus does not require logarithms, so find $n$ by trial and improvement (or by listing values with the calculator's table or a spreadsheet). At 7% p.a. compounding annually, $1.07^{10} \approx 1.967$ and $1.07^{11} \approx 2.105$ , so an investment first doubles after 11 years. Always give the first whole number of periods that meets the target.

The effect of the rate and the compounding frequency

A higher interest rate increases the total amount, and the effect grows over time because interest compounds on interest.
A longer time increases the amount geometrically, not linearly: the amount grows faster and faster.
More frequent compounding (at the same nominal rate) gives a slightly larger amount, because interest starts earning interest sooner. For $10 000 at 6% p.a. for 3 years:

Compounding	Total amount
Annually	$11 910.16
Quarterly	$11 956.18
Monthly	$11 966.81
Daily	$11 972.00

The gain from more frequent compounding is real but small, and it gets smaller as the frequency increases. That is why comparing offers with different compounding frequencies is best done with the effective annual rate.

Worked examples: amount, principal, rate and time

Total amount and interest

$12 000 is invested at 4.8% p.a. compounding monthly for 5 years. Find the total amount and the interest.

Convert: $i = \frac{0.048}{12} = 0.004$ , $n = 5 \times 12 = 60$ .
Formula: $A = 12\,000(1.004)^{60} \approx 15\,247.69$ .
Interest: $15\,247.69 - 12\,000 = 3247.69$ .

Marker's note: write $i$ and $n$ as a separate line before substituting. It earns method marks and prevents the most common error (using the annual rate).

Finding the principal

What amount invested now at 5.2% p.a. compounding quarterly grows to $20 000 in 6 years?

Convert: $i = \frac{0.052}{4} = 0.013$ , $n = 24$ .
Rearrange: $P = \frac{20\,000}{1.013^{24}} \approx 14\,669.09$ .
Answer: $14 669.09.

Marker's note: check by substituting back: $14\,669.09 \times 1.013^{24} \approx 20\,000$ .

Finding the interest rate

$8000 grows to $9500 in 4 years, compounding annually. Find the interest rate.

Set up: $9500 = 8000(1 + i)^4$ , so $(1 + i)^4 = 1.1875$ .
Root: $1 + i = 1.1875^{1/4} \approx 1.04390$ .
Answer: $i \approx 0.0439$ , about 4.39% p.a.

Marker's note: keep at least five significant figures in the intermediate root before converting to a percentage.

Finding the number of periods

$5000 is invested at 4.8% p.a. compounding monthly. When does it first exceed $7000?

Model: $A = 5000(1.004)^n$ .
Trial and improvement: $n = 80$ : 6881.25. $n = 84$ : 6992.0. $n = 85$ : 7019.98.
Answer: After 85 months.

Marker's note: show the two values either side of the target; they justify your answer without logarithms.

Common traps

Using the annual rate as $i$: For monthly compounding, $i$ is the annual rate divided by 12.
Using years as $n$: $n$ counts compounding periods: 5 years of monthly compounding is $n = 60$ .
Confusing $r$ and $i$: In the QCAA notation $r = 1 + i$ is the multiplier; $i$ is the rate. A rate of 0.4% per month means $i = 0.004$ and $r = 1.004$ .
Giving the amount when the interest was asked for: Interest is $A - P$ .
Treating compound interest as linear: Interest is not the same every year; it grows because it is calculated on a growing balance.
Rounding too early: Keep full calculator values until the final answer, then round money to the nearest cent.

Exam technique

Start every compound interest answer with a line listing $P$ , $i$ and $n$ (and $A$ if known). Use the recurrence when the question asks you to "show" the amount after the first few periods, and the formula for anything further ahead. For "how long" questions, show the calculator values for the period before and after the target. Round money to the nearest cent and rates to the precision asked.

Note

Compound interest means your interest earns interest. If you put money in a bank account, each month the bank adds a little bit, and next month it adds a bit on the new, bigger total. So the money grows faster and faster, like a snowball rolling downhill. The formula just says: start amount, times the growth factor, as many times as there are months (or quarters, or years). A loan works the same way, except it is your debt that snowballs.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

$3000 is invested at 5% p.a. compounding annually for 4 years. Find (a) the total amount and (b) the total interest earned.

Show worked solution →

(a) $i = 0.05$ , $n = 4$ : $A = 3000(1.05)^4 = 3000 \times 1.21550625 = 3646.52$ . The total amount is $3646.52. (1 mark)

(b) Interest $= A - P = 3646.52 - 3000 = 646.52$ , so $646.52. (1 mark)

foundation2 marks

An investment is modelled by

A_{n+1} = 1.004A_n

A_0 = 12\,000

, where

n

is in months. (a) State the interest rate per month and per annum. (b) Find

A_3

, showing each step.

Show worked solution →

(a) $r = 1.004 = 1 + i$ , so $i = 0.004$ : $0.4\%$ per month, which is $0.4 \times 12 = 4.8\%$ p.a. compounding monthly. (1 mark)

(b) $A_1 = 1.004 \times 12\,000 = 12\,048.00$ ; $A_2 = 1.004 \times 12\,048.00 = 12\,096.19$ ; $A_3 = 1.004 \times 12\,096.19 = 12\,144.58$ . (1 mark, with each step shown.)

foundation2 marks

$25 000 is invested at 3.6% p.a. compounding quarterly for 3 years. Find

i

n

and the total amount.

Show worked solution →

$i = \frac{0.036}{4} = 0.009$ per quarter; $n = 3 \times 4 = 12$ quarters. (1 mark)

$A = 25\,000(1.009)^{12} \approx 27\,837.74$ . The total amount is $27 837.74. (1 mark)

core2 marks

How much must be invested now at 5.4% p.a. compounding monthly to have $15 000 in 4 years? Give the answer to the nearest cent.

Show worked solution →

$i = \frac{0.054}{12} = 0.0045$ , $n = 48$ . Rearrange $A = P(1 + i)^n$ :

P = \frac{A}{(1 + i)^n} = \frac{15\,000}{1.0045^{48}} \approx 12\,091.89.

(1 mark for the rearrangement and values, 1 mark for the answer.) $12 091.89 must be invested.

core2 marks

$10 000 grew to $14 000 in 5 years with interest compounding annually. Find the annual interest rate, as a percentage to two decimal places.

Show worked solution →

$14\,000 = 10\,000(1 + i)^5$ , so $(1 + i)^5 = 1.4$ and

1 + i = 1.4^{1/5} \approx 1.069610.

(1 mark) $i \approx 0.0696$ , so the rate is $6.96\%$ p.a. (1 mark) On a scientific calculator, use the $x$ th root key (or $1.4 \wedge (1 \div 5)$ ).

core2 marks

$5000 is invested at 4.8% p.a. compounding monthly. After how many whole months will the total amount first exceed $7000?

Show worked solution →

$A = 5000(1.004)^n$ . Try values of $n$ (a table on the calculator helps): $5000 \times 1.004^{84} \approx 6992.0$ and $5000 \times 1.004^{85} \approx 7019.98$ . (1 mark)

The amount first exceeds $7000 after 85 months (7 years and 1 month). (1 mark) Logarithms are not required: systematic trial and improvement is enough.

exam3 marks

A $5000 compound interest loan charges 9.6% p.a. compounding monthly and is repaid in a single payment after 2 years. (a) Find the amount owed after 2 years. (b) Find the total interest. (c) Explain why the interest in the second year is greater than in the first year.

Show worked solution →

(a) $i = 0.008$ , $n = 24$ : $A = 5000(1.008)^{24} \approx 6053.73$ . (1 mark)

(b) Interest $= 6053.73 - 5000 = 1053.73$ . (1 mark)

(c) Interest is charged on the balance, which includes the interest already added. In year 2 the balance is larger (after one year it is $5000 \times 1.008^{12} \approx 5501.70$ ), so each month's interest is larger than in year 1. The first year's interest is about $501.70 and the second year's is about $552.03. (1 mark)

exam3 marks

Ava has $10 000 to invest for 5 years. Option A pays 5.8% p.a. compounding annually. Option B pays 5.6% p.a. compounding monthly. (a) Find the total amount for each option. (b) Which option should she choose? (c) Explain why a higher compounding frequency did not make Option B better.

Show worked solution →

(a) A: $10\,000(1.058)^5 \approx 13\,256.48$ . B: $10\,000\left(1 + \frac{0.056}{12}\right)^{60} \approx 13\,222.68$ . (1 mark)

(b) Option A, by about $33.80. (1 mark)

(c) Monthly compounding increases the effective rate of Option B above 5.6%, but only to about 5.75% p.a., which is still less than Option A's 5.8%. More frequent compounding helps, but it cannot make up for a lower nominal rate unless the difference is very small. (1 mark)

Practise this

Flashcards General Mathematics deck

Sources & how we know this

General Mathematics General senior syllabus 2025 — QCAA