Financial Mathematics

Reducing balance loans and annuities modelled with recurrence relations and geometric series - repayments, amount owing, time to repay, future value of an annuity, drawing down a lump sum, and how changing the rate or payment affects the outcome.

1What you need to know
2The key idea: build it from a recurrence
3Reducing balance loans
4Annuities
5Common mistakes
6Exam tips
7Links to other topics

1.What you need to know

Financial mathematics was worth 7 marks on NESA's sample paper for the 2024 syllabus. In the new syllabus it covers just two things, reducing balance loans and annuities, and both are geometric series problems. The 2024 syllabus does not list compound interest or depreciation as separate dot points; they appear as growth and decay contexts for arithmetic and geometric sequences in MA12-SS.

Reducing balance loans: a loan charged compound interest each period and paid off with equal periodic repayments. Find the repayment, the amount owing after $n$ repayments, the total paid and the time to repay.
Annuities: regular equal deposits into an account earning compound interest (future value), or a lump sum from which regular equal withdrawals are made.
Investigate how the interest rate, payment size and term affect a loan or investment, using a formula, a table or a spreadsheet.

Payments in this course are made at the end of each period.

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2.The key idea: build it from a recurrence

Every loan and annuity problem starts the same way. Let $r$ be the interest rate per period as a decimal (6% p.a. paid monthly gives $r = 0.005$ ) and let $A_{n}$ be the balance after $n$ periods. Write the first few balances without simplifying, spot the geometric series, then sum it with

S_{n} = \frac{a ( R ^{n} - 1 )}{R - 1}

where $R = 1 + r$ is the common ratio. Questions very often ask you to "show that" a formula holds, and the marks are for exactly this build-up.

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3.Reducing balance loans

A loan of $P$ is charged interest at rate $r$ per period, then a repayment $M$ is made at the end of each period. The recurrence is

A_{n + 1} = A_{n} (1 + r) - M, A_{0} = P

Writing out the first few balances:

A_{1} A_{2} A_{3} = P (1 + r) - M = P (1 + r)^{2} - M [1 + (1 + r)] = P (1 + r)^{3} - M [1 + (1 + r) + (1 + r)^{2}]

The bracket is a GP with first term 1, ratio $1 + r$ and $n$ terms, so

A_{n} = P (1 + r)^{n} - M \cdot \frac{( 1 + r ) ^{n} - 1}{r}

The loan is paid off when $A_{n} = 0$ . Solving for $M$ gives the repayment for a loan over $n$ periods:

M = \frac{P r ( 1 + r ) ^{n}}{( 1 + r ) ^{n} - 1}

You can learn this, but it is safer to know how to get it from $A_{n} = 0$ .

Worked example 1

Ravi borrows $18 000 to buy a car. Interest is 7.2% p.a., compounded monthly, and he makes equal monthly repayments over 4 years.

(a) Find the monthly repayment. Here $r = \frac{0.072}{12} = 0.006$ and $n = 48$ . Setting $A_{48} = 0$ :

M = \frac{18 000 \times 0.006 \times 1.00 6 ^{48}}{1.00 6 ^{48} - 1} \approx $432.70

(b) How much interest does he pay? Total paid is $48 \times 432.70 = $20 769.60$ , so the interest is $$20 769.60 - $18 000 = $2769.60$ .

(c) How much does he still owe after one year? (using the rounded repayment $M = 432.70$ )

A_{12} = 18000 (1.006)^{12} - 432.70 \times \frac{1.00 6 ^{12} - 1}{0.006} \approx $13 972.41

After a quarter of the term he has paid off only about $4000 of the $18 000. Early repayments are mostly interest; the principal falls faster near the end of the loan.

How long to repay?

When $M$ is fixed and $n$ is unknown, set $A_{n} = 0$ and solve for $(1 + r)^{n}$ , then use logarithms. The answer is usually not a whole number: round up, because the last repayment is a smaller one. The loan can only be repaid if $M$ is more than the first period's interest, $P r$ ; otherwise the balance never falls.

Worked example 2

A loan of $12 000 is charged 1% interest per month, and $350 is repaid at the end of each month. How many repayments are needed?

12000 (1.01)^{n} 12000 (1.01)^{n} 35000 1.0 1^{n} = 350 \cdot \frac{1.0 1 ^{n} - 1}{0.01} = 35000 (1.0 1^{n} - 1) = 23000 (1.01)^{n} = \frac{35}{23}

n = \frac{ln ( 35/23 )}{ln 1.01} \approx 42.2

So 43 repayments are needed (42 full ones and a smaller final one).

Changing the rate or the repayment

The syllabus asks you to investigate how the interest rate and the repayment affect the time to repay. For the same $12 000 loan, the number of monthly repayments needed is:

Repayment	0.75% p.m.	1% p.m.	1.25% p.m.
$350	40	43	46
$400	35	36	38
$500	27	28	29

Two patterns to be able to explain: a higher repayment shortens the loan and reduces total interest; a higher interest rate lengthens it, and the effect of the rate is much larger when the repayment is only just above the interest charged each month.

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4.Annuities

An annuity is a sequence of equal payments at equal intervals, earning compound interest. There are two kinds.

Future value: saving up

Deposit $M$ at the end of each period into an account earning $r$ per period. The first deposit earns interest for $n - 1$ periods, the second for $n - 2$ , and the last for none. Reading the balance just after the $n$ th deposit from the last deposit back:

F V = M + M (1 + r) + \dots + M (1 + r)^{n - 1} = M \cdot \frac{( 1 + r ) ^{n} - 1}{r}

The future value is all the deposits plus all the interest they have earned.

If the account also starts with an initial deposit $P$ , that lump sum grows on its own as well, so

F V = P (1 + r)^{n} + M \cdot \frac{( 1 + r ) ^{n} - 1}{r}

The syllabus asks you to examine how changing the initial amount, the regular payment, the rate and the term changes the final value; this formula (or a spreadsheet built from the recurrence $A_{n + 1} = A_{n} (1 + r) + M$ ) is how you do it.

Drawing down a lump sum

A lump sum $P$ is invested now, and a withdrawal $W$ is made at the end of each period after interest is added (like a retirement income). This is a reducing balance loan in reverse: the account plays the bank. The balance after $n$ withdrawals is

A_{n} = P (1 + r)^{n} - W \cdot \frac{( 1 + r ) ^{n} - 1}{r}

and the money lasts exactly $n$ periods when $A_{n} = 0$ . Solving $A_{n} = 0$ for $P$ gives the lump sum needed to fund $n$ withdrawals:

P = W \cdot \frac{1 - ( 1 + r ) ^{- n}}{r}

Treat this as a result you derive from $A_{n} = 0$ , not a formula to memorise.

Tables of interest factors

Older HSC papers sometimes gave a table of future value or present value interest factors. The 2024 syllabus does not mention these tables, so expect to use the geometric series formula instead.

Worked example 3

Mei deposits $350 at the end of every quarter into an account earning 6% p.a., compounded quarterly, for 5 years. Find the value of the account just after the last deposit, and how much of it is interest.

Here $r = \frac{0.06}{4} = 0.015$ and $n = 20$ .

F V = 350 \cdot \frac{1.01 5 ^{20} - 1}{0.015} \approx $8093.28

She deposited $20 \times 350 = $7000$ , so the interest earned is about $1093.28.

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5.Common mistakes

Using the annual rate with monthly periods ( $r = 0.072$ instead of $0.006$ ), or the number of years instead of the number of periods.
Getting the number of terms in the geometric series wrong. Write out $A_{1}$ , $A_{2}$ , $A_{3}$ and count.
Rounding the time to repay down instead of up.
Rounding $M$ before using it again in a later part. Use the value the question gives or your unrounded calculator value, and say which.
Mixing up saving up to a target (future value, deposits added) with drawing down a lump sum (withdrawals subtracted).
Assuming payments are at the start of each period. In this course they are at the end unless a question clearly says otherwise; if it does, write out the recurrence again rather than reusing a formula.

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6.Exam tips

Show that a formula for $A_{n}$ : write $A_{1}$ , $A_{2}$ and $A_{3}$ in unsimplified form, identify the geometric series (first term, ratio, number of terms), then sum it. Each step usually carries a mark.
Find the repayment: set $A_{n} = 0$ and rearrange; show the substitution with the period rate and number of periods.
How many repayments: solve with logs and explain the rounding in a sentence.
Money answers to the nearest cent unless the question says otherwise, with a $ sign.
In "investigate" or "explain" parts, refer to numbers from your working (for example, total interest under each option) to support the comparison.

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7.Links to other topics

MA12-SS Sequences and series: the geometric series sum is the whole engine of this topic, and compound interest itself is a GP.
MA11-EL Exponential and logarithmic functions: logarithms solve for the number of payments.
MA11-FN Working with functions: rearranging formulas to make $M$ , $P$ or $(1 + r)^{n}$ the subject.

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Financial Mathematics

Contents

1.What you need to know

2.The key idea: build it from a recurrence

3.Reducing balance loans

Worked example 1

How long to repay?

Worked example 2

Changing the rate or the repayment

4.Annuities

Future value: saving up

Drawing down a lump sum

Tables of interest factors

Worked example 3

5.Common mistakes

6.Exam tips

7.Links to other topics

Test yourself