Reducing balance loans (and their link to annuities): VCE General Mathematics Unit 3 Recursion and financial modelling

Syllabus dot point

“Use a first-order linear recurrence relation and an amortisation table to model and analyse a reducing balance loan, and use technology to find the repayment, balance, term, final payment and total interest, including the impact of a change in interest rate or repayment”

18 min readUpdated 23 Sep 2026

Quick answer

A reducing balance loan follows $V_{n+1} = RV_n - d$ , where $V_0$ is the amount borrowed: interest on the balance is added and the repayment subtracted. In an amortisation table, interest $=$ previous balance $\times$ rate, principal reduction $=$ repayment $-$ interest. Use the finance solver ( $PV$ positive, $PMT$ negative for a borrower) for repayments, balances, terms and final repayments; total interest $=$ total repaid $-$ amount borrowed; after any change, the current balance becomes the new starting value.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

A reducing balance loan (a car loan, personal loan or home loan) is repaid with equal regular repayments. Each period interest is charged on the balance still owing, then the repayment is subtracted, so the balance falls ("reduces") until it reaches zero. VCAA wants you to model this with a first-order recurrence relation, investigate it step by step with an amortisation table, and use the CAS finance solver to find the repayment, the balance at any time, the number of repayments, the final repayment, the total interest, and the effect of changing the interest rate or the repayment.

This is examined heavily. The 2023 papers contained four loan questions (two multiple-choice and a five-part Examination 2 question) and 2025 Examination 2 opened its finance section with an amortisation table. The report statistics quoted below show how often small slips cost marks.

(The same mathematics describes an annuity, where you invest a lump sum and receive the payments. That has its own page; the connection is explained at the end.)

The answer

The recurrence relation

If $V_n$ is the balance after $n$ repayments, $R = 1 + \frac{r}{100}$ is the growth factor for the interest rate $r\%$ per period, and $d$ is the repayment:

V_0 = \text{amount borrowed}, \qquad V_{n+1} = RV_n - d.

For $20 000 at $9\%$ per annum compounding monthly ( $0.75\%$ per month) with repayments of $913.70: $V_0 = 20\,000$ , $V_{n+1} = 1.0075V_n - 913.70$ . After one month, $V_1 = 20\,150 - 913.70 = 19\,236.30$ : the balance fell by $763.70, because $150 of the repayment covered the interest.

To convert an annual rate: monthly divide by 12, quarterly by 4, fortnightly by 26, weekly by 52. So $7.8\%$ per annum compounding weekly gives $R = 1 + \frac{7.8}{5200} = 1.0015$ . Going the other way, $(R - 1) \times 52 \times 100\%$ recovers the annual rate, which is exactly what 2023 Examination 2 Question 7a asked students to show (only $32\%$ did).

The amortisation table

An amortisation table lays out each repayment:

Payment	Payment ($)	Interest ($)	Principal reduction ($)	Balance ($)
0	0.00	0.00	0.00	350 000.00
1	2255.05	1750.00	505.05	349 494.95
2	2255.05	1747.47	507.58	348 987.37
3	2255.05	1744.94	510.11	348 477.26

($350 000 at 6% per annum compounding monthly, 25 years.)

Interest $=$ previous balance $\times$ rate per period ( $350\,000 \times 0.005 = 1750$ ).
Principal reduction $=$ payment $-$ interest.
Balance $=$ previous balance $-$ principal reduction.

Early in a long loan most of each repayment is interest (here $1750 of $2255.05); the principal falls slowly at first and faster later. The interest column falls each row because it is calculated on a lower balance, which is exactly the explanation the 2025 examiners wanted for Declan's loan.

Key fact

Reducing balance loan: $V_0$ is the amount borrowed, $V_{n+1} = RV_n - d$ with $R = 1 + \frac{r}{100}$ for the rate per repayment period. Amortisation row: interest $=$ previous balance $\times \frac{r}{100}$ , principal reduction $=$ repayment $-$ interest, new balance $=$ previous balance $-$ principal reduction. Total interest $=$ total repaid $-$ amount borrowed. If $d$ equals the interest, the loan is interest-only (balance constant); if $d = 0$ , the sequence is geometric.

The finance solver

The finance solver links $N$ (number of repayments), $I(\%)$ (annual rate), $PV$ , $PMT$ , $FV$ and $PpY$ , $CpY$ (repayments and compounding periods per year). For a loan, from the borrower's point of view:

Field	Loan entry
$PV$	amount borrowed, positive (money received)
$PMT$	repayment, negative (money paid out)
$FV$	0 when fully repaid; the balance owing (negative) part-way through

Standard uses:

Repayment for a given term: fill $N$ , $I(\%)$ , $PV$ , $FV = 0$ ; solve $PMT$ .
Balance after $k$ repayments: set $N = k$ and solve $FV$ (the balance is the size of the answer).
Number of repayments: fill $I(\%)$ , $PV$ , $PMT$ , $FV = 0$ ; solve $N$ . A decimal answer means a smaller final repayment.
Interest rate: fill $N$ , $PV$ , $PMT$ , $FV$ ; solve $I(\%)$ .

The final repayment

If $N$ is not a whole number, the loan needs a final repayment smaller than the others. Find the balance after the last full repayment (set $N$ to the whole number below and solve for $FV$ ), then add one period's interest. For Timmy's loan the balance after 299 repayments is $2605.65 and the final repayment is $2605.65 \times 1.00325 = 2614.12$ . Choosing $2605.65 (forgetting the last month's interest) was the most popular wrong answer in 2023.

Total interest

\text{total interest} = \text{total of all repayments} - \text{amount borrowed}.

A $450 000 loan at $6.12\%$ per annum over 30 years with repayments of $2732.79 costs about $360 \times 2732.79 - 450\,000 = 533\,804.40$ dollars in interest, more than the amount borrowed.

Changing the rate or the repayment

Higher repayments shorten the loan and reduce total interest. Paying $3000 instead of $2732.79 on the $450 000 loan repays it in 285 months instead of 360 and saves roughly $129 800 in interest.
A higher interest rate means either higher repayments (to keep the end date) or a longer term (if the repayment is unchanged), and more interest either way.
Part-way changes use the balance at the moment of change as the new $PV$ . After 5 years the $450 000 loan's balance is $419 360.99; if the rate rises to $6.6\%$ , keeping the end date needs $2857.82 a month, while keeping $2732.79 extends the loan by about 39 months.

Interest-only loans

If each repayment equals the interest for the period, the balance never falls: an interest-only loan (the loan version of a perpetuity). For $60 000 at $0.15\%$ per week, the interest-only repayment is $0.0015 \times 60\,000 = 90$ dollars per week.

Loans and annuities are the same model

An annuity (invest a lump sum, receive regular payments until it runs out) follows the same recurrence $V_{n+1} = RV_n - d$ and the same amortisation table. Only the point of view changes: with an annuity you are the lender, so on the finance solver $PV$ is negative (money you invest) and $PMT$ is positive (money you receive). See the annuities page for annuity questions.

Worked examples: recurrence, table, solver and changes

Two recursive steps

$V_0 = 20\,000$ , $V_{n+1} = 1.0075V_n - 913.70$ . Find $V_2$ .

$V_1 = 1.0075 \times 20\,000 - 913.70 = 19\,236.30$ .

$V_2 = 1.0075 \times 19\,236.30 - 913.70 = 19\,380.57 - 913.70 = 18\,466.87$ .

Marker's note: keep each balance to the cent before using it in the next step.

The next row of an amortisation table

Declan's balance after payment 2 is $824 443.59, the rate is $0.35\%$ per month and the payment is $15 730.88.

Interest: $824\,443.59 \times 0.0035 = 2885.55$ .
Principal reduction: $15\,730.88 - 2885.55 = 12\,845.33$ .
Balance: $824\,443.59 - 12\,845.33 = 811\,598.26$ .

Marker's note: only $40\%$ of students got all three values in 2025; round each to the cent before the next step.

Repayment and interest with the solver

$450 000 at $6.12\%$ per annum compounding monthly over 30 years.

Inputs: $N = 360$ , $I(\%) = 6.12$ , $PV = 450\,000$ , $FV = 0$ , $PpY = CpY = 12$ .
Repayment: $PMT = -2732.79$ .
Interest: $360 \times 2732.79 - 450\,000 \approx 533\,804$ .

Marker's note: list the inputs; they earn method marks.

A rate rise part-way through

After 5 years the balance is $419 360.99 and the rate rises to $6.6\%$ .

Keep the end date. $N = 300$ , $I(\%) = 6.6$ , $PV = 419\,360.99$ , $FV = 0$ : $PMT = -2857.82$ .

Keep the repayment. $PMT = -2732.79$ , solve $N \approx 338.7$ : about 39 extra months.

Marker's note: the balance at the change is the new $PV$ ; never restart from $450 000.

Common traps

Using the annual rate in $R$: $6\%$ per annum monthly is $R = 1.005$ .
Subtracting the whole repayment from the balance: Only the principal reduction comes off.
Forgetting the last period's interest in a final repayment: Final repayment $=$ balance after the last full repayment $\times R$ .
Wrong signs in the solver: For a loan, $PV$ positive and $PMT$ negative (or the reverse consistently); mixed-up signs give nonsense.
Restarting from the original amount after a change: Use the balance at the change.
Rounding too early: Keep full precision in intermediate steps; round money to the cent when writing each table value.
Thinking Examination 1 is technology-free: Both General Mathematics examinations allow CAS, so the finance solver can be used in multiple-choice questions.

Exam technique

Write the solver inputs as a list before every solver answer. For recurrence questions, write the per-period rate and $R$ first. For amortisation rows, write interest, principal reduction and balance in that order. When a question changes something part-way, draw a quick timeline marking the balance at the change. Check answers for sense: balances fall, final repayments are smaller than regular ones, and higher rates cost more.

Note

When you borrow money, the bank adds interest to what you owe each month, then takes off your repayment. Early on, most of your repayment just covers the interest, so the debt shrinks slowly; later, more of each repayment actually pays down the debt. Paying a bit more each month shortens the loan a lot and saves heaps of interest, while a rate rise does the opposite. An amortisation table is just a month-by-month record of this: interest added, repayment made, debt left.

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.

2023 VCAA-style1 mark

Timmy took out a reducing balance loan of $500 000, with interest calculated monthly. The balance of the loan after

n

months is modelled by

T_0 = 500\,000

T_{n+1} = 1.00325T_n - 2611.65

. A final repayment that will fully repay the loan, to the nearest cent, is A. $2605.65 B. $2609.18 C. $2611.65 D. $2614.12 E. $2615.81

Show worked answer →

Use the finance solver (or the recurrence) to find how many full repayments are made: solving for $N$ with $I(\%) = 0.325 \times 12 = 3.9$ , $PV = 500\,000$ , $PMT = -2611.65$ , $FV = 0$ , $PpY = CpY = 12$ gives $N \approx 299.998$ , so 299 full repayments.

After 299 repayments the balance is $2605.65 (set $N = 299$ and solve for $FV$ ). One more month of interest is added before the final payment:

2605.65 \times 1.00325 \approx 2614.12.

The answer is D. Option A is the balance before the last month's interest, the most common wrong choice. Only $39\%$ of students were correct.

Source: VCAA 2023 General Mathematics Examination 1, Question 22, and the 2023 examination report.

2023 VCAA-style1 mark

Tavi took out a loan of $20 000, with interest compounding quarterly, and makes quarterly repayments of $653.65. The balance at the end of each quarter of the first year is $20 000, $19 527.56, $19 050.83, $18 569.79, $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 →

Recover the quarterly growth factor from one step: $20\,000R - 653.65 = 19\,527.56$ , so

R = \frac{19\,527.56 + 653.65}{20\,000} = 1.009\,060\,5.

The quarterly rate is $0.906\,05\%$ , a nominal $3.6242\%$ per annum. The effective rate is $\left(1.009\,060\,5^4 - 1\right) \times 100\% \approx 3.674\%$ (or use the formula sheet with $r = 3.6242$ and $n = 4$ ). The answer is D ( $37\%$ correct).

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

2025 VCAA-style4 marks

Declan has a reducing balance loan with interest calculated monthly at 4.2% per annum and monthly repayments. Rows of the amortisation table: payment 0: balance $850 000.00; payment 1: payment $15 730.88, interest $2975.00, principal reduction $12 755.88, balance $837 244.12; payment 2: payment $15 730.88, interest $2930.35, principal reduction $12 800.53, balance $824 443.59. (a) What amount did Declan borrow? (b) Why is the interest for payment 2 lower than for payment 1? (c) Complete the row for payment 3. (d) The last payment is $15 730.71. How many payments of $15 730.88 are made before it?

Show worked answer →

(a) $850 000 (the balance at payment 0). ( $93\%$ correct.)

(b) Interest is calculated on the balance, and the balance has been reduced by payment 1, so the interest for payment 2 is calculated on a lower (reducing) balance. ( $65\%$ correct.)

(c) Monthly rate $4.2 \div 12 = 0.35\%$ . Interest $= 824\,443.59 \times 0.0035 = 2885.55$ . Principal reduction $= 15\,730.88 - 2885.55 = 12\,845.33$ . Balance $= 824\,443.59 - 12\,845.33 = 811\,598.26$ . ( $40\%$ correct.)

(d) Finance solver: $I(\%) = 4.2$ , $PV = 850\,000$ , $PMT = -15\,730.88$ , $FV = 0$ , $PpY = CpY = 12$ gives $N \approx 59.9999$ . So there are 59 full payments of $15 730.88 before the final (slightly smaller) payment. ( $41\%$ correct.)

Source: VCAA 2025 General Mathematics Examination 2, Question 7, and the 2025 examination report.

2023 VCAA-style5 marks

Arthur takes out a loan of $60 000 with interest compounding weekly. The balance after

n

weeks is

V_0 = 60\,000

V_{n+1} = 1.0015V_n - d

. (a) Show that the interest rate is 7.8% per annum. (b) Find

d

if (i) Arthur makes interest-only repayments; (ii) he repays the loan in full in five years (to the nearest cent). (c) If

d = 300

for the first year, what value of

d

fully repays the loan with exactly three more years of repayments? (d) For what value of

d

is the sequence geometric?

Show worked answer →

(a) $(1.0015 - 1) \times 52 \times 100\% = 0.15\% \times 52 = 7.8\%$ per annum. ( $32\%$ correct; the working had to be shown because the answer was given.)

(b)(i) Interest only: $d = 0.0015 \times 60\,000 = 90$ . ( $47\%$ correct.)

(b)(ii) Finance solver: $N = 260$ , $I(\%) = 7.8$ , $PV = 60\,000$ , $FV = 0$ , $PpY = CpY = 52$ : $PMT = -278.86$ , so $d = 278.86$ . ( $40\%$ correct.)

(c) After 52 weeks at $300 the balance is about $48 651.67; repaying that over 156 weeks gives $d = 350.01$ . ( $32\%$ correct.)

(d) $d = 0$ (no repayments: pure compound growth, a geometric sequence). Only $11\%$ of students answered this correctly.

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

A $20 000 loan at 9% per annum, compounding monthly, is repaid with monthly repayments of $913.70. (a) Write a recurrence relation for the balance

V_n

after

n

months. (b) Find the balance after one month, showing the calculation.

Show worked solution →

(a) Monthly rate $9 \div 12 = 0.75\%$ , so $V_0 = 20\,000$ , $V_{n+1} = 1.0075V_n - 913.70$ . (1 mark)

(b) $V_1 = 1.0075 \times 20\,000 - 913.70 = 20\,150 - 913.70 = 19\,236.30$ . (1 mark) The balance fell by $763.70, not by the whole $913.70, because $150 of the payment went to interest.

foundation2 marks

For the $20 000 loan in the previous question, repaid in 24 payments of $913.70, find (a) the total amount repaid and (b) the total interest paid (ignore any adjustment to the final payment).

Show worked solution →

(a) $24 \times 913.70 = 21\,928.80$ . (1 mark)

(b) $21\,928.80 - 20\,000 = 1928.80$ . (1 mark)

foundation3 marks

A $350 000 loan charges 6% per annum compounding monthly, with repayments of $2255.05 per month. Complete the first three rows of the amortisation table.

Show worked solution →

Monthly rate $0.5\%$ . Each row: interest $=$ previous balance $\times 0.005$ ; principal reduction $=$ payment $-$ interest; balance $=$ previous balance $-$ principal reduction.

Payment	Payment ($)	Interest ($)	Principal reduction ($)	Balance ($)
0	0.00	0.00	0.00	350 000.00
1	2255.05	1750.00	505.05	349 494.95
2	2255.05	1747.47	507.58	348 987.37
3	2255.05	1744.94	510.11	348 477.26

(3 marks: one per correct row.)

core3 marks

A $450 000 home loan charges 6.12% per annum compounding monthly and is to be repaid over 30 years. (a) Find the monthly repayment. (b) Find the total interest paid (using the rounded repayment for all 360 payments). (c) Write the finance solver inputs you used.

Show worked solution →

(c) $N = 360$ , $I(\%) = 6.12$ , $PV = 450\,000$ , $PMT = ?$ , $FV = 0$ , $PpY = CpY = 12$ . (1 mark)

(a) $PMT = -2732.79$ , so $2732.79 per month. (1 mark) (The negative sign shows money paid out by the borrower.)

(b) $360 \times 2732.79 - 450\,000 = 983\,804.40 - 450\,000 = 533\,804.40$ . (1 mark)

core2 marks

A borrower owes $60 000 on a loan at 7.8% per annum compounding weekly (0.15% per week). What weekly repayment would keep the balance constant, and what is this type of loan called?

Show worked solution →

The repayment must equal the weekly interest: $0.0015 \times 60\,000 = 90$ , so $90 per week. (1 mark)

This is an interest-only loan: the principal is never reduced (the loan equivalent of a perpetuity). (1 mark)

core2 marks

For the $450 000 loan at 6.12% (repayments $2732.79 per month), the borrower instead pays $3000 per month. (a) How many payments are needed? (b) Roughly how much interest is saved?

Show worked solution →

(a) Finance solver with $PMT = -3000$ , $FV = 0$ gives $N \approx 284.7$ : 284 full payments and a smaller final payment (285 payments, about 23.75 years). (1 mark)

(b) Total paid $\approx 284 \times 3000 + 2036.64 = 854\,036.64$ , so interest $\approx 404\,036.64$ , compared with about $533 804 at the original repayment: a saving of roughly $129 800. (1 mark)

exam4 marks

A $450 000 loan at 6.12% per annum compounding monthly has repayments of $2732.79 (30-year term). After 5 years the rate rises to 6.6% per annum. (a) Find the balance after 5 years. (b) Find the new repayment that keeps the original end date. (c) If instead the repayment stays at $2732.79, how much longer does the loan take (to the nearest month)?

Show worked solution →

(a) $N = 60$ , $I(\%) = 6.12$ , $PV = 450\,000$ , $PMT = -2732.79$ : $FV \approx -419\,360.99$ , a balance of $419 360.99. (1 mark)

(b) $N = 300$ , $I(\%) = 6.6$ , $PV = 419\,360.99$ , $FV = 0$ : $PMT \approx -2857.82$ , so $2857.82 per month. (1 mark)

(c) $I(\%) = 6.6$ , $PV = 419\,360.99$ , $PMT = -2732.79$ , $FV = 0$ : $N \approx 338.7$ months, compared with the 300 months remaining. The loan takes about 39 months (3 years and 3 months) longer. (2 marks)

exam2 marks

The balance of a loan is modelled by

V_0 = 80\,000

V_{n+1} = RV_n - d

, with

n

in months. The balance after one month is $79 600 and after two months $79 198. Find

R

d

and the annual interest rate.

Show worked solution →

$79\,600 = 80\,000R - d$ and $79\,198 = 79\,600R - d$ . Subtracting: $402 = 400R$ , so $R = 1.005$ . (1 mark)

$d = 80\,000 \times 1.005 - 79\,600 = 80\,400 - 79\,600 = 800$ . The monthly rate is $0.5\%$ , so $6\%$ per annum compounding monthly. (1 mark)

Practise this

Sources & how we know this

VCE General Mathematics Examinations 1 and 2 (2023) and examination reports — VCAA (2023)
VCE General Mathematics Examination 2 (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