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

An annuity is what a single lump sum is not: instead of investing one amount and letting it grow, you add the same amount every period. The key insight that unlocks the whole topic is that each deposit is just an ordinary compound-interest problem, and they only differ in how long each one has been sitting in the account. The very first deposit compounds the longest; the very last earns nothing because it lands at the end. Add up all those separately-grown deposits and you get a geometric series, whose sum is the future-value formula. Everything else is rearranging that one equation.

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. (The other convention, an annuity due, pays at the start of each period; it earns one extra period of interest, so its future value is the ordinary-annuity value times $(1 + r)$. Assume an ordinary annuity unless told otherwise.)

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.

Watching the future value build up, stage by stage

To see why this is a sum and not a single power, build the future value one deposit at a time. Take four annual deposits of $2000 into an account paying $6\%$ per annum, valued just after the fourth deposit. Each deposit is drawn as a bar showing what it is worth at the end: the earliest deposit has compounded longest, so it is the tallest, and the running total appears under each panel.

Stage 1, the first payment. Payment $1$ is made at the end of year $1$, so it has $3$ full years left to compound before the end of year $4$. It grows to $2000 \times (1.06)^3 = 2382.03$, i.e. $2382.03.

Stage 2, add the second payment. Payment $2$ arrives a year later, so it compounds for only $2$ years: $2000 \times (1.06)^2 = 2247.20$, i.e. $2247.20. The running total of the first two payments is $4629.23.

Stage 3, add the third payment. Payment $3$ compounds for just $1$ year, adding $2000 \times 1.06 = 2120.00$, i.e. $2120.00. The running total climbs to $6749.23.

Stage 4, add the last payment. Payment $4$ is made at the very end of year $4$, so it earns no interest and adds exactly $2000.00. The four bars sum to $8749.23, which is precisely what the formula gives: $2000 \cdot \dfrac{(1.06)^4 - 1}{0.06} = 2000 \times 4.374616 = 8749.23$, i.e. $8749.23.

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 $(1 + r)$. The safest defence against these off-by-one variations is to track the first and last deposits explicitly, as the small-case check in the worked example does.

How exam questions ask about annuities

Each wording is one of the three rearrangements of the same formula:

• "Find the balance / value just after the $n$th deposit." Straight future-value formula $A = M\dfrac{(1 + r)^n - 1}{r}$. Convert the rate first.
• "How much must each deposit be to reach $X?" Solve for the payment, $M = \dfrac{Ar}{(1 + r)^n - 1}$.
• "How many deposits / years are needed to reach $X?" Solve for $n$ with logs, $n = \dfrac{\ln(Ar/M + 1)}{\ln(1 + r)}$, and round up to the next whole deposit.
• "How much interest was earned?" Future value minus total deposited, $A - Mn$.
• "Show that the balance is a geometric series" or "derive the formula". Write each deposit's grown value, identify $a = M$ and ratio $(1 + r)$ over $n$ terms, then apply the sum formula. Markers reward the explicit derivation.
• "The deposits are made at the start of each period." An annuity due: compute the ordinary-annuity value and multiply by $(1 + r)$.