Simple and compound interest: VCE General Mathematics Unit 3 Recursion and financial modelling

Syllabus dot point

“The concepts of simple and compound interest; use of a recurrence relation to model and analyse (numerically and graphically) a compound interest investment or loan; the difference between nominal and effective interest rates and the use of effective interest rates to compare investment returns and the cost of loans; and the future value of a compound interest investment or loan after n compounding periods”

13 min readUpdated 23 Sep 2026

Quick answer

Simple interest adds the same amount every period, $d = \frac{r}{100} \times V_0$ , so $V_{n+1} = V_n + d$ and $V_n = V_0 + nd$ (linear growth). Compound interest multiplies by $R = 1 + \frac{r}{100}$ every period, where $r$ is the rate per compounding period, so $V_{n+1} = RV_n$ and $V_n = R^n V_0$ (geometric growth). Divide the annual rate by the number of compounding periods per year, and count $n$ in periods. The effective annual rate, $\left(1 + \frac{r}{100n}\right)^n - 1$ , is the true yearly growth and is the fair way to compare investments or loans with different compounding.

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From interest to the rest of the topic
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What this dot point is asking

The Study Design topic "Compound interest investments and loans" lists:

"the concepts of simple and compound interest";
"use of a recurrence relation to model and analyse (numerically and graphically) a compound interest investment or loan, including the use of a recurrence relation to determine the value of the compound interest loan or investment after $n$ compounding period for an initial sequence from first principles";
"the difference between nominal and effective interest rates and the use of effective interest rates to compare investment returns and the cost of loans when interest is paid or charged, for example, daily, monthly, quarterly";
"the future value of a compound interest investment or loan after $n$ compounding periods and its use to solve practical problems".

So you need both models as recurrence relations and rules, the rate per compounding period, the future value, and effective rates. These basics are where many finance marks are lost: in 2025 Examination 1, $32\%$ used compound interest for a simple interest question (Question 17) and only $41\%$ used an effective rate correctly (Question 23); in 2023 Examination 1 Question 23 only $37\%$ found an effective rate.

The answer

Simple interest: linear growth

Simple interest is calculated on the original principal only, so the interest is the same every period. For a principal $V_0$ at $r\%$ per annum:

\text{interest per year} = \frac{r}{100} \times V_0, \qquad I = \frac{V_0 r t}{100} \text{ for } t \text{ years}.

As a recurrence, with $d$ the interest per period:

V_0 = \text{principal}, \qquad V_{n+1} = V_n + d, \qquad V_n = V_0 + nd.

The balances form an arithmetic sequence and the graph of balance against time is a straight line. For $4000 at $4\%$ per annum, $d = 160$ , so after three years the balance is $4000 + 3 \times 160 = 4480$ dollars.

Compound interest: geometric growth

Compound interest is calculated on the current balance, so interest earns interest. If the rate is $r\%$ per compounding period:

V_0 = \text{principal}, \qquad V_{n+1} = RV_n, \qquad R = 1 + \frac{r}{100}, \qquad V_n = R^n V_0.

The balances form a geometric sequence and the graph curves upwards: each year's increase is larger than the last.

Rates per compounding period

Interest rates are quoted per annum, but interest is often compounded more often. Divide the annual rate by the number of compounding periods per year, and count $n$ in those periods:

Compounding	Periods per year	Rate per period for 4.8% p.a.	$R$	$n$ for 3 years
yearly	1	4.8%	1.048	3
quarterly	4	1.2%	1.012	12
monthly	12	0.4%	1.004	36
fortnightly	26	0.1846%	1.001846	78
weekly	52	0.0923%	1.000923	156

Going the other way, a recurrence $V_{n+1} = 1.0015V_n$ with weekly compounding means $0.15\%$ per week, and $0.15\% \times 52 = 7.8\%$ per annum (2023 Examination 2 Question 7a).

Compound interest loans

A loan with no repayments grows in exactly the same way: $5000 borrowed at $9.6\%$ per annum compounding monthly grows to $5000 \times 1.008^{24} = 6053.73$ dollars after two years. With regular repayments the recurrence becomes $V_{n+1} = RV_n - d$ (a reducing balance loan), which is why $d = 0$ gives a geometric sequence in 2023 Examination 2 Question 7d.

Nominal and effective rates

The quoted annual rate is the nominal rate. When interest compounds more than once a year, interest earns interest within the year, so the balance grows by more than the nominal rate. The effective annual rate is the actual percentage growth over one year:

r_{\text{effective}} = \left(\left(1 + \frac{r}{100n}\right)^n - 1\right) \times 100\%,

where $r$ is the nominal rate and $n$ is the number of compounding periods per year (this is on the formula sheet). On CAS, TI-Nspire has eff(nominal, n) and nom(effective, n); ClassPad has the same conversions in its Financial application.

For $4.8\%$ per annum: compounding monthly gives $1.004^{12} - 1 = 4.91\%$ effective.

Why effective rates matter. Two investments with different compounding cannot be compared by their nominal rates. The effective rate puts both on the same yearly basis:

Offer	Effective rate
4.6% p.a. compounding yearly	4.600%
4.5% p.a. compounding monthly	4.594%
4.52% p.a. compounding daily	4.623%
4.55% p.a. compounding quarterly	4.628%

For an investment choose the highest effective rate; for a loan the lowest.

Using an effective rate. Because the effective rate is the true annual growth, a balance after a whole number of years is simply $V_0 \times (1 + r_{\text{eff}})^{\text{years}}$ (2025 Examination 1 Question 23).

Finance solver for compound interest

Compound interest is a finance solver problem with $PMT = 0$ : for $12 000 at $4.8\%$ per annum compounding monthly for three years, enter $N = 36$ , $I\% = 4.8$ , $PV = -12\,000$ , $PMT = 0$ , $P/Y = C/Y = 12$ , and solve for $FV = 13\,854.63$ . The solver also finds $N$ (time to reach a target) or $I\%$ (rate needed).

Formula

Simple: $V_{n+1} = V_n + d$ , $V_n = V_0 + nd$ , $d = \frac{r}{100}V_0$ . Compound: $V_{n+1} = RV_n$ , $V_n = R^nV_0$ , $R = 1 + \frac{r}{100}$ with $r$ per compounding period. Effective rate $= \left(1 + \frac{r}{100n}\right)^n - 1$ (nominal $r$ , $n$ periods a year). Interest earned $=$ final value $-$ principal.

Worked examples: models, rates and comparisons

Which expression is simple interest? (2025 Examination 1 Question 17)

Step 1: simple interest is $0.04 \times 4000 = 160$ dollars every year.
Step 2: three years add $3 \times 160$ : balance $= 4000 + 3(0.04 \times 4000) = 4480$ . Answer: D.
Step 3 (check the distractor): $4000 \times 1.04^3 = 4499.46$ is compound interest, $19.46 more because interest earns interest.

Compound interest from first principles

$12 000 is invested at $4.8\%$ per annum compounding monthly.

Step 1: the recurrence: Monthly rate $0.4\%$ , so $V_0 = 12\,000$ , $V_{n+1} = 1.004V_n$ .
Step 2: iterate: $V_1 = 12\,048$ ; $V_2 = 1.004 \times 12\,048 = 12\,096.19$ ; $V_3 = 1.004 \times 12\,096.192 = 12\,144.58$ dollars.
Step 3: the rule: After three years ( $n = 36$ ): $V_{36} = 1.004^{36} \times 12\,000 = 13\,854.63$ dollars, so the interest earned is $1854.63.
Step 4: effective rate: $1.004^{12} - 1 = 4.91\%$ .

An effective rate from a loan graph (2023 Examination 1 Question 23)

Step 1: one quarter's interest: The balance falls from $20 000 to $19 527.56 after a repayment of $653.65: interest $= 19\,527.56 - 20\,000 + 653.65 = 181.21$ .
Step 2: nominal rate: $\frac{181.21}{20\,000} = 0.906\,05\%$ per quarter, so $3.6242\%$ per annum.
Step 3: effective rate: $(1.009\,060\,5)^4 - 1 = 3.67\%$ . Answer: D.

Check with the graph: $19\,527.56 \times 1.009\,060\,5 - 653.65 = 19\,050.84$ , matching the second point ($19 050.83) to within a cent.

Interest from an effective rate (2025 Examination 1 Question 23)

Step 1: the effective rate is the yearly growth: $5000 \times 1.0451^5 = 6233.89$ dollars.

Step 2: interest $= 1233.89$ , closest to $1234. Answer: B.

How long to double?

$10 000 at $5\%$ per annum compounding quarterly: solve $1.0125^n = 2$ . CAS gives $n = 55.8$ , so the balance first doubles after 56 quarters (14 years). Check: $1.0125^{55} = 1.980$ and $1.0125^{56} = 2.005$ .

What rate is needed?

$8000 is to grow to $10 000 in four years with monthly compounding. Solve $8000\left(1 + \frac{r}{1200}\right)^{48} = 10\,000$ , or Finance Solver with $N = 48$ , $PV = -8000$ , $PMT = 0$ , $FV = 10\,000$ , $P/Y = C/Y = 12$ : $r = 5.59\%$ per annum.

Common traps

Compounding simple interest: Simple interest adds the same amount each year: $4000 + 3(0.04 \times 4000)$ , not $4000 \times 1.04^3$ .
Using the annual rate every period: For monthly compounding at $4.8\%$ p.a., $R = 1.004$ , not $1.048$ , and $n$ counts months.
Treating an effective rate as nominal: An effective rate is already the yearly growth. Entering 4.51% as the nominal rate compounding fortnightly overstates the interest (2025 Examination 1 Question 23, option D).
Reporting the nominal rate when the effective rate is asked for: The effective rate is higher whenever interest compounds more than once a year (3.67%, not 3.62%, in 2023).
Comparing nominal rates with different compounding: Convert all to effective rates first.
Not showing working in a "show that": Write $(1.0015 - 1) \times 52 \times 100 = 7.8$ ; a CAS solve line was not accepted in 2023.
Rounding a time down: "First doubles" or "first exceeds" needs the next whole period.

Exam technique

Before calculating, write the three things that set up every compound interest question: the rate per period, $R$ , and $n$ in periods. For "which is simple or compound" multiple choice, look for a power (compound) versus a multiple of a constant amount (simple). For effective-rate questions, write the formula-sheet expression with the numbers in it, even if you use CAS to evaluate it.

From interest to the rest of the topic

Every model in Recursion and financial modelling is a first-order linear recurrence $V_{n+1} = RV_n + d$ . Simple interest is the case $R = 1$ (add a constant); compound interest is $d = 0$ (multiply by a constant). Adding regular deposits gives an annuity investment, subtracting repayments gives a reducing balance loan or an annuity, and repayments equal to the interest give a perpetuity or interest-only loan. Reducing balance depreciation is the same geometric model with $R$ less than 1.

Note

Simple interest pays the same amount every year, worked out on the money you first put in, so your balance goes up in a straight line. Compound interest pays interest on the whole balance, including earlier interest, so the balance grows faster and faster. Banks quote a yearly rate, but if they add interest monthly you use one-twelfth of it each month. The effective rate is how much your money really grows in a year, and it is the fair way to compare two accounts.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2025 VCAA Exam 1 Q17Past paper1 mark

Dani invests $4000 for three years. The account earns simple interest at 4% per annum. The balance in the account after three years can be calculated using A.

4000 \times 1.04^3

3(1.04 \times 4000)

4000 + (0.04 \times 4000)^3

4000 + 3(0.04 \times 4000)

Show worked answer →

Simple interest is the same amount every year, calculated on the original principal:

\text{interest each year} = 0.04 \times 4000 = 160, \qquad \text{balance} = 4000 + 3(0.04 \times 4000) = 4480.

The answer is D ( $53\%$ correct). $32\%$ chose A, $4000 \times 1.04^3 = 4499.46$ , which is compound interest: it multiplies by 1.04 each year, so interest is earned on interest.

Source: VCAA 2025 General Mathematics Examination 1, Question 17, and the 2025 examination report.

2025 VCAA Exam 1 Q23Past paper1 mark

Virat invested $5000 into an account that earned interest compounding fortnightly. The effective annual interest rate for Virat's investment was 4.51%. Assume that there are exactly 26 fortnights in one year. After five years, the amount of interest earned by Virat was closest to A. $1128 B. $1234 C. $1262 D. $1264

Show worked answer →

The effective rate already includes the fortnightly compounding, so the balance grows by a factor of 1.0451 each whole year:

5000 \times 1.0451^5 = 6233.89, \qquad \text{interest} = 6233.89 - 5000 = 1233.89.

The report's method gives the same result: the nominal rate is $4.415\%$ (CAS: nom(4.51, 26)), then Finance Solver with $N = 130$ , $I\% = 4.415$ , $PV = -5000$ , $PMT = 0$ , $P/Y = C/Y = 26$ gives $FV = 6233.89$ .

The answer is B ( $41\%$ correct). $32\%$ chose D, from treating 4.51% as the nominal rate compounding fortnightly ($1263.52 of interest); A is simple interest at 4.51% ($1127.50).

Source: VCAA 2025 General Mathematics Examination 1, Question 23, and the 2025 examination report.

2023 VCAA Exam 1 Q23Past paper1 mark

Tavi took out a loan of $20 000, with interest compounding quarterly. She makes quarterly repayments of $653.65. The graph below represents the balance in dollars of Tavi's loan at the end of each quarter of the first year of the loan. [Points (quarter, balance): (0, 20 000), (1, 19 527.56), (2, 19 050.83), (3, 18 569.79), (4, 18 084.39).] The effective interest rate for the first year of Tavi's loan is closest to A. 3.62% B. 3.65% C. 3.66% D. 3.67% E. 3.68%

Show worked answer →

Step 1: interest in the first quarter: $20\,000 + \text{interest} - 653.65 = 19\,527.56$ , so the interest is $181.21.
Step 2: rates: Quarterly rate $= \frac{181.21}{20\,000} = 0.906\,05\%$ ; nominal annual rate $= 4 \times 0.906\,05\% = 3.6242\%$ .
Step 3: effective rate

r_{\text{effective}} = \left(1 + \frac{3.6242}{400}\right)^4 - 1 = 0.036\,74 \approx 3.67\%.

The answer is D ( $37\%$ correct). $22\%$ chose A, the nominal rate; the effective rate is always a little higher when interest compounds more than once a year.

Source: VCAA 2023 General Mathematics Examination 1, Question 23, and the 2023 examination report.

2023 VCAA Exam 2 Q7a,dPast paper2 marks

Arthur takes out a new loan of $60 000 to pay for an overseas holiday. Interest on this loan compounds weekly. The balance of the loan, in dollars, after

n

weeks,

V_n

, can be determined using a recurrence relation of the form

V_0 = 60\,000

V_{n+1} = 1.0015V_n - d

. a. Show that the interest rate for this loan is 7.8% per annum. d. For what value of

d

does the recurrence relation generate a geometric sequence?

Show worked answer →

a. The weekly growth factor is 1.0015, so the weekly rate is $1.0015 - 1 = 0.0015$ (0.15%). With 52 weeks in a year:

(1.0015 - 1) \times 52 \times 100 = 7.8\% \text{ per annum}.

Only $32\%$ were correct: every step had to be shown, and a CAS "solve" line was not accepted.

d. $d = 0$ . Then $V_{n+1} = 1.0015V_n$ : each term is the previous one multiplied by a constant, a geometric sequence (a compound interest loan with no repayments). Only $11\%$ were correct.

Source: VCAA 2023 General Mathematics Examination 2, Question 7, and the 2023 examination report.

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

$2500 is invested at 3.5% per annum simple interest. (a) Write a recurrence relation for the balance

V_n

after

n

years. (b) Find the balance after four years.

Show worked solution →

(a) Interest each year $= 0.035 \times 2500 = 87.50$ , so $V_0 = 2500$ , $V_{n+1} = V_n + 87.5$ . (1 mark)

(b) $V_4 = 2500 + 4 \times 87.5 = 2850$ dollars. (1 mark)

foundation1 mark

An investment of $3000 earning simple interest grows to $3360 in three years. Find the annual interest rate.

Show worked solution →

Interest $= 360$ over 3 years, so $120 a year. Rate $= \frac{120}{3000} \times 100\% = 4\%$ per annum. (1 mark)

core2 marks

$20 000 is invested at 3.6% per annum, compounding monthly. (a) Write a recurrence relation for the balance

V_n

after

n

months. (b) Showing recursive calculations, find the balance after three months, to the nearest cent.

Show worked solution →

(a) Monthly rate $\frac{3.6}{12} = 0.3\%$ , so $V_0 = 20\,000$ , $V_{n+1} = 1.003V_n$ . (1 mark)

(b) $V_1 = 1.003 \times 20\,000 = 20\,060$ ; $V_2 = 1.003 \times 20\,060 = 20\,120.18$ ; $V_3 = 1.003 \times 20\,120.18 = 20\,180.54$ dollars. (1 mark)

core2 marks

$6000 is invested for five years at 3% per annum, compounding monthly. Find (a) the balance at the end (b) the interest earned, both to the nearest cent.

Show worked solution →

(a) $R = 1 + \frac{3}{1200} = 1.0025$ and $n = 60$ : $6000 \times 1.0025^{60} = 6969.70$ dollars. (1 mark)

(b) $6969.70 - 6000 = 969.70$ dollars. (1 mark)

core2 marks

Account A pays 6.2% per annum compounding quarterly. Account B pays 6.1% per annum compounding monthly. Use effective interest rates to decide which account gives the better return.

Show worked solution →

A: $\left(1 + \frac{0.062}{4}\right)^4 - 1 = 6.35\%$ . B: $\left(1 + \frac{0.061}{12}\right)^{12} - 1 = 6.27\%$ . (1 mark)

Account A has the higher effective rate, so it gives the better return, even though B compounds more often. (1 mark)

exam2 marks

A borrower can choose a loan at 12.5% per annum compounding daily (365 days a year) or 12.8% per annum compounding monthly. Which loan costs less? Justify with effective rates to two decimal places.

Show worked solution →

Daily: $\left(1 + \frac{0.125}{365}\right)^{365} - 1 = 13.31\%$ . Monthly: $\left(1 + \frac{0.128}{12}\right)^{12} - 1 = 13.58\%$ . (1 mark)

The 12.5% loan compounding daily has the lower effective rate, so it costs less. (1 mark)

exam2 marks

An investment earns 4.2% per annum compounding monthly. After how many whole months will it first be worth at least 1.5 times its starting value?

Show worked solution →

Solve $1.0035^n = 1.5$ (CAS, or Finance Solver with $PV = -1$ , $FV = 1.5$ , $PMT = 0$ , $I\% = 4.2$ , $P/Y = C/Y = 12$ ): $n = 116.05$ . (1 mark)

After 116 months it is just short of 1.5 times, so it takes 117 months (9 years and 9 months). (1 mark)

Practise this

Sources & how we know this

VCE General Mathematics Examination 1 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 2 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 1 (2025) and examination report — VCAA (2025)
VCE Mathematics Study Design: General Mathematics (from 2023) — VCAA
VCE General Mathematics: examination specifications, past examinations and reports — VCAA