Derive and use the future value formula for an annuity to find the accumulated value of a series of equal regular contributions

What this dot point is asking

NESA wants you to model regular equal contributions to an investment, derive the future value of an ordinary annuity using a geometric series, apply it to standard savings problems, and rearrange it to solve for the required contribution.

The answer

What is an annuity

An annuity is a sequence of equal payments made at regular intervals. In the ordinary annuity model used in Maths Advanced, each payment is made at the end of a compounding period, and interest is credited at the same rate per period that compounding occurs.

Let $M$ be the payment per period, $r$ the per-period interest rate (as a decimal), and $n$ the number of payments. We want the balance $A$ just after the $n$th payment.

Deriving the formula

Track when each payment is made and how many full periods it earns interest before time $n$.

• Payment $1$ is made at the end of period $1$ and earns interest for $n - 1$ periods. Its value at time $n$ is $M(1 + r)^{n - 1}$.
• Payment $2$ is made at the end of period $2$ and earns interest for $n - 2$ periods. Its value at time $n$ is $M(1 + r)^{n - 2}$.
• The $n$th payment is made at time $n$ and earns no interest. Its value is $M$.

The total balance is the sum

This is a geometric series with first term $M$, common ratio $(1 + r)$ and $n$ terms. So

This is the future value of an ordinary annuity.

Rearranging for other unknowns

Solve for $M$ when the target balance $A$ is given:

Solve for $n$ (with $A$, $M$, $r$ given):

There is no closed-form solution for $r$; in the exam, $r$ is always given.

Sanity checks

• Total deposited is $M n$. The future value $A$ exceeds $M n$ because of interest.
• If $r = 0$, the formula simplifies to $A = M n$ (interpreting the indeterminate form $0 / 0$ in the limit).
• Doubling $n$ more than doubles $A$, because later contributions sit in the account longer and earlier contributions compound longer.

A common variation: deposit-then-credit

Some questions credit interest at the start of each period instead of after, or count the balance just before the next deposit. The number of compounding periods for each deposit changes by $1$. Read the question carefully and either reuse the geometric-series derivation or multiply $A$ by an extra factor of $(1 + r)$.

Worked examples

Direct future value

Deposit \500$at the end of each quarter into an account paying$8%$per annum compounded quarterly for$5$ years.

$r = \frac{0.08}{4} = 0.02$, $n = 20$, $M = 500$.

A = 500 \cdot \frac{(1.02)^{20} - 1}{0.02} = 500 \cdot \frac{1.48595 - 1}{0.02} = 500 \cdot 24.2974 \approx \12148.69$.

Total deposited: 500 \cdot 20 = \10000$. Interest earned: about$\$2148.69$.

Finding the required payment

You want \100000$in$20$years. Account pays$6%$ per annum compounded monthly. Monthly deposit?

$r = \frac{0.06}{12} = 0.005$, $n = 240$, $A = 100000$.

$(1.005)^{240} \approx 3.31020$. Denominator: $2.31020$.

M = \frac{100000 \cdot 0.005}{2.31020} = \frac{500}{2.31020} \approx \216.43$.

Finding the time

Deposit \1000$at the end of each year at$5%$per annum compounded annually. How many years to reach$\$25000$?

$r = 0.05$, $M = 1000$, $A = 25000$.

$\frac{A r}{M} + 1 = \frac{25000 \cdot 0.05}{1000} + 1 = 1.25 + 1 = 2.25$.

$n = \frac{\ln 2.25}{\ln 1.05} = \frac{0.81093}{0.04879} \approx 16.62$ years.

So $17$ full annual deposits are needed to reach or exceed \25000$.

Building the series by hand

For a small case, list the contributions and check the formula. Three deposits of \100$at the end of each year at$10%$ per annum:

Payment $1$ grows for $2$ years: $100(1.1)^2 = 121$.

Payment $2$ grows for $1$ year: $100(1.1) = 110$.

Payment $3$ is the deposit itself: $100$.

Total: 121 + 110 + 100 = \331$. Formula check:$100 \cdot \frac{(1.1)^3 - 1}{0.1} = 100 \cdot 3.31 = \$331$.

Common traps

Wrong number of compounding periods per payment. In an ordinary annuity the last payment earns zero interest. Off-by-one errors are common; always check with a small case.

Using the annual rate with monthly payments. Convert to the per-period rate first.

Confusing total deposits with the future value. Total deposited is $M n$. Future value $A$ is larger by the interest earned. Some questions ask for the interest, which is $A - M n$.

Wrong direction in the rearranged formula. When solving for $M$, the formula is $M = \frac{A r}{(1 + r)^n - 1}$, not $M = \frac{A}{(1 + r)^n}$ (that is a single lump-sum discount).

Mixing ordinary annuity and annuity due. Annuity due payments are made at the start of each period and earn one extra period of interest, multiplying the future value by $(1 + r)$. The default in Maths Advanced is the ordinary annuity unless stated otherwise.

In one sentence

The future value of an ordinary annuity is the geometric sum $A = M \cdot \frac{(1 + r)^n - 1}{r}$, which rearranges to give the required payment or, with logarithms, the required number of periods.