Annuities and future value: deriving the formula and applying it to regular savings

Syllabus dot point

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

HSCMaths AdvancedYear 12: Financial Mathematics15 min readUpdated 20 Jun 2026

Quick answer

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.

Jump to a section

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

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.

Note

Think of pushing children on a queue of swings, one after another. The first child you push has the whole afternoon to keep swinging higher, so by home time she is soaring; the last child you push right before you leave barely moves at all. An annuity is the money version of this: you put in the same amount every month, but the earliest deposits have the most time to grow with interest, while the newest deposit has earned almost nothing yet. To find the total in the account, you cannot just multiply, because each deposit has grown by a different amount. Instead you add up every separately-grown deposit, and that sum is exactly the future-value formula.

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

A = M + M(1 + r) + M(1 + r)^2 + \cdots + M(1 + r)^{n - 1}.

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

A = M \cdot \frac{(1 + r)^n - 1}{(1 + r) - 1} = M \cdot \frac{(1 + r)^n - 1}{r}.

This is the future value of an ordinary annuity.

Key fact

The future value of an ordinary annuity (payments at the end of each period) is the geometric sum $A = M \cdot \dfrac{(1 + r)^n - 1}{r}$ , where $M$ is the payment, $r$ the per-period rate and $n$ the number of payments. Rearrange it for the required payment, $M = \dfrac{A r}{(1 + r)^n - 1}$ , or for the number of periods, $n = \dfrac{\ln(Ar/M + 1)}{\ln(1 + r)}$ . There is no closed form for $r$ , so the exam always gives it. Convert the nominal annual rate to a per-period rate before substituting.

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:

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

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

n = \frac{\ln \left( \frac{A r}{M} + 1 \right)}{\ln(1 + r)}.

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 $(1 + r)$ .

Worked example

Four calculations showing how to apply the future value formula, solve for the required deposit, find the number of periods, and verify the formula from first principles.

Find the future value of a quarterly annuity

Convert to per-period quantities: Every annuity problem begins with identifying the three inputs: per-period rate $r$ , number of periods $n$ , and payment $M$ . Rates quoted annually must be divided by the number of compounding periods per year.

For the first example: $500 deposited 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, \qquad n = 5 \times 4 = 20, \qquad M = 500.

Apply the future value formula: Substitute into $A = M \cdot \frac{(1+r)^n - 1}{r}$ . Compute the compound factor first to keep arithmetic clean.

(1.02)^{20} \approx 1.48595, \qquad (1.02)^{20} - 1 \approx 0.48595.

A = 500 \cdot \frac{0.48595}{0.02} = 500 \cdot 24.2974 \approx 12148.69.

So the future value is about $12148.69. Total deposited: $500 \times 20 = 10000$ , i.e. $10000. Interest earned: about $2148.69.

Solve for the required payment

You want $100000 in $20$ years. Account pays $6\%$ per annum compounded monthly. Rearrange the formula for $M$ : $M = \frac{A r}{(1+r)^n - 1}$ .

r = \frac{0.06}{12} = 0.005, \qquad n = 20 \times 12 = 240, \qquad A = 100000.

(1.005)^{240} \approx 3.31020, \qquad (1.005)^{240} - 1 \approx 2.31020.

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

So each monthly deposit must be about $216.43.

Find the number of periods using logarithms

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

Set $A = 25000$ , $M = 1000$ , $r = 0.05$ and solve for $n$ using the logarithm formula:

\frac{A r}{M} + 1 = \frac{25000 \times 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 \text{ years.}

Because $n$ must be a whole number of deposits, $17$ full annual deposits are needed to reach or exceed $25000.

Verify the formula from first principles (small case)

Track each payment individually to confirm the formula is correct. Three deposits of $100 at the end of each year at $10\%$ per annum:

Payment $1$ is made at the end of year $1$ and compounds for $2$ years: $100 \times (1.1)^2 = 121$ .

Payment $2$ is made at the end of year $2$ and compounds for $1$ year: $100 \times 1.1 = 110$ .

Payment $3$ is made at the end of year $3$ and earns no interest: $100$ .

Total: $121 + 110 + 100 = 331$ , i.e. $331. Formula check: $100 \cdot \frac{(1.1)^3 - 1}{0.1} = 100 \times 3.31 = 331$ , i.e. $331. The formula matches.

Final answers: future value $\approx$ $12148.69; required monthly deposit $\approx$ $216.43; time needed: $17$ full years.

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.

Exam technique

Write $r$ , $n$ and $M$ as a labelled first line; markers award method marks for the correct per-period rate and period count before any arithmetic. Compute the compound factor $(1 + r)^n$ to several decimal places and round only the final figure, to cents. State which quantity you are solving for so the rearrangement is clear, and use the future-value form, not a single lump-sum discount. For "how many deposits" questions, take logs and round up. If a derivation is asked for, show the stack of grown deposits and name $a = M$ and the ratio $(1 + r)$ ; if the deposits are made at the start of the period, multiply the answer by $(1 + r)$ and say why.

Exam-style questions

Questions in the style of NESA 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.

HSC-style4 marks

Maria deposits $200 at the end of each month into an account paying

6\%

per annum compounded monthly. Find the balance just after her 60th deposit.

Show worked answer →

Per-period rate: $r = \frac{0.06}{12} = 0.005$ . Payment: $M = 200$ . Number of payments: $n = 60$ .

The future value of an ordinary annuity is

$A = M \cdot \frac{(1 + r)^n - 1}{r} = 200 \cdot \frac{(1.005)^{60} - 1}{0.005}$ .

$(1.005)^{60} \approx 1.34885$ , so $(1.005)^{60} - 1 \approx 0.34885$ .

$A \approx 200 \cdot \frac{0.34885}{0.005} = 200 \cdot 69.770 \approx 13954.01$ , i.e. $13954.01.

Markers reward the per-period rate, the right formula, an accurate compound factor, and a final answer to cents.

HSC-style4 marks

Jia wants to save $50000 over

10

years by depositing an equal amount at the end of each month into an account paying

4.8\%

per annum compounded monthly. How much must each deposit be?

Show worked answer →

$r = \frac{0.048}{12} = 0.004$ , $n = 120$ , $A = 50000$ .

Rearrange the future-value formula for $M$ :

$M = \frac{A \cdot r}{(1 + r)^n - 1} = \frac{50000 \cdot 0.004}{(1.004)^{120} - 1}$ .

$(1.004)^{120} \approx 1.614528$ , so the denominator is $0.614528$ .

$M = \frac{200}{0.614528} \approx 325.45$ , i.e. $325.45.

Markers expect the correct per-period rate, the rearrangement for $M$ , an accurate compound factor, and a final answer to cents.

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 saver deposits $300 at the end of each month into an account paying

6\%

per annum compounded monthly. Find the balance just after the 24th deposit.

Show worked solution →

Set up the per-period quantities. The annual rate is compounded monthly, so

r = \frac{0.06}{12} = 0.005, \qquad n = 24, \qquad M = 300.

Apply the future-value formula.

A = M \cdot \frac{(1 + r)^n - 1}{r} = 300 \cdot \frac{(1.005)^{24} - 1}{0.005}.

With $(1.005)^{24} \approx 1.12716$ , the numerator is $0.12716$ , so

A = 300 \cdot \frac{0.12716}{0.005} = 300 \times 25.43196 \approx 7629.59,

i.e. $7629.59.

Marker's note: one mark for the per-period rate $r = 0.005$ with $n = 24$ , one for substituting into $A = M\frac{(1+r)^n - 1}{r}$ and reaching $7629.59 to the nearest cent. Using the annual rate $0.06$ or $n = 2$ (years) loses the method mark.

foundation3 marks

An investor deposits $1500 at the end of each year for

6

years into an account paying

7\%

per annum compounded annually. Find the future value and hence the total interest earned.

Show worked solution →

Identify the inputs. Payments are annual and the rate is compounded annually, so

r = 0.07, \qquad n = 6, \qquad M = 1500.

Compute the future value.

A = 1500 \cdot \frac{(1.07)^6 - 1}{0.07}.

Since $(1.07)^6 \approx 1.50073$ , the numerator is $0.50073$ , giving

A = 1500 \cdot \frac{0.50073}{0.07} = 1500 \times 7.15329 \approx 10729.94,

i.e. $10729.94.

Interest earned is future value minus total deposited. The total put in is $Mn = 1500 \times 6 = 9000$ , so

\text{interest} = 10729.94 - 9000 = 1729.94,

i.e. $1729.94.

Marker's note: one mark for the future value $10729.94, one for computing total deposits $Mn = 9000$ , one for the interest $1729.94. Quoting $A$ itself as the interest is the classic slip.

core3 marks

A saver deposits $400 at the end of each month into an account paying

6\%

per annum compounded monthly. Find the smallest number of deposits needed for the balance to first reach $20000.

Show worked solution →

Set up and rearrange for $n$ . Here $r = \frac{0.06}{12} = 0.005$ , $M = 400$ and the target is $A = 20000$ . Rearranging the future-value formula,

n = \frac{\ln\!\left(\dfrac{A r}{M} + 1\right)}{\ln(1 + r)}.

Substitute. First the bracket:

\frac{A r}{M} + 1 = \frac{20000 \times 0.005}{400} + 1 = \frac{100}{400} + 1 = 1.25.

Then

n = \frac{\ln 1.25}{\ln 1.005} = \frac{0.22314}{0.0049875} \approx 44.74.

Round up to a whole deposit. The balance passes $20000 partway through the 45th month, so $45$ deposits are needed.

Marker's note: one mark for the rearrangement and the bracket value $1.25$ , one for $n \approx 44.74$ from the logarithms, one for rounding up to $45$ . Rounding to $44$ (or truncating) leaves the balance short of $20000 and loses the final mark.

core4 marks

A savings plan makes three annual deposits of $2500 at

8\%

per annum compounded annually. The table below records, for each deposit, the number of full years it compounds before the end of year

3

and its grown value at that time. Deposit 1: compounds for

2

years, grown value $2916.00. Deposit 2: compounds for

1

year, grown value $2700.00. Deposit 3: compounds for

0

years, grown value $2500.00. (a) Explain why deposit 1 compounds for only

2

years, not

3

. (b) Use the table to find the future value, and confirm it with the annuity formula.

Show worked solution →

Part (a): read the timing from the ordinary-annuity model. Each deposit is made at the end of its year. Deposit 1 lands at the end of year $1$ , so between then and the end of year $3$ only $3 - 1 = 2$ full years pass. That is why it compounds for $2$ years, and why the final deposit (end of year $3$ ) earns nothing.

Part (b): add the grown values in the table.

A = 2916.00 + 2700.00 + 2500.00 = 8116.00,

i.e. $8116.00.

Confirm with the formula using $r = 0.08$ , $M = 2500$ , $n = 3$ :

A = 2500 \cdot \frac{(1.08)^3 - 1}{0.08} = 2500 \cdot \frac{1.259712 - 1}{0.08} = 2500 \times 3.2464 = 8116.00,

i.e. $8116.00. The table sum and the formula agree.

Marker's note: one mark for the end-of-year timing explanation in (a), one for summing the table to $8116.00, one for a correct formula substitution, one for the matching $8116.00. A candidate who says deposit 1 compounds for $3$ years has misread the ordinary-annuity convention and loses part (a).

core3 marks

How much must be deposited at the end of each month into an account paying

4.8\%

per annum compounded monthly to accumulate $80000 after

15

years?

Show worked solution →

Set up and rearrange for the payment. The target and inputs are

r = \frac{0.048}{12} = 0.004, \qquad n = 15 \times 12 = 180, \qquad A = 80000,

and the required payment is

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

Substitute. With $(1.004)^{180} \approx 2.05148$ , the denominator is $1.05148$ , so

M = \frac{80000 \times 0.004}{1.05148} = \frac{320}{1.05148} \approx 304.33,

i.e. each monthly deposit must be about $304.33.

Marker's note: one mark for $r = 0.004$ and $n = 180$ , one for the rearrangement $M = \frac{Ar}{(1+r)^n - 1}$ , one for $304.33 to the nearest cent. Dividing by $(1+r)^n$ instead of $(1+r)^n - 1$ is a single-lump-sum discount and is wrong here.

exam4 marks

A worker pays $250 into a superannuation fund at the start of each month (an annuity due) for

40

years. The fund earns

6\%

per annum compounded monthly. (a) Explain why the future value of an annuity due is the ordinary-annuity value multiplied by

(1 + r)

. (b) Find the balance at the end of the

40

years, to the nearest dollar.

Show worked solution →

Part (a): each payment earns one extra period of interest. In an annuity due every deposit is made at the start of its period rather than the end, so each one sits in the account one full period longer than in the ordinary model. Multiplying every term of the ordinary-annuity sum by $(1 + r)$ accounts for that single extra period, so

A_{\text{due}} = A_{\text{ordinary}} \times (1 + r).

Part (b): compute the ordinary value, then adjust. Here

r = \frac{0.06}{12} = 0.005, \qquad n = 40 \times 12 = 480, \qquad M = 250.

With $(1.005)^{480} \approx 10.95745$ , the numerator is $9.95745$ , so the ordinary value is

A_{\text{ordinary}} = 250 \cdot \frac{9.95745}{0.005} = 250 \times 1991.4907 \approx 497872.68.

Multiply by $(1 + r)$ for the annuity due:

A_{\text{due}} = 497872.68 \times 1.005 \approx 500362,

i.e. about $500362.

Marker's note: one mark for the "one extra period" reasoning in (a), one for $r = 0.005$ and $n = 480$ , one for the ordinary value $497872.68, one for multiplying by $1.005$ to reach $500362. Forgetting the $(1 + r)$ factor caps the answer at the ordinary value.

exam5 marks

A graduate deposits $600 at the end of each quarter for

5

years into a fund paying

8\%

per annum compounded quarterly, then stops depositing but leaves the balance invested at the same rate for a further

3

years. (a) Show that the balance immediately after the final deposit is about $14578. (b) Hence find the balance at the end of the full

8

years, to the nearest dollar.

Show worked solution →

Part (a): the deposit phase is an ordinary annuity. Quarterly compounding gives

r = \frac{0.08}{4} = 0.02, \qquad n = 5 \times 4 = 20, \qquad M = 600.

Apply the future-value formula:

A = 600 \cdot \frac{(1.02)^{20} - 1}{0.02}.

With $(1.02)^{20} \approx 1.48595$ , the numerator is $0.48595$ , so

A = 600 \cdot \frac{0.48595}{0.02} = 600 \times 24.2974 \approx 14578.42,

which is about $14578, as required.

Part (b): no more deposits, so the balance grows as a single lump sum. For the next $3$ years there are $3 \times 4 = 12$ more quarterly compoundings and no annuity payments, so

B = 14578.42 \times (1.02)^{12}.

With $(1.02)^{12} \approx 1.26824$ ,

B = 14578.42 \times 1.26824 \approx 18488.96,

i.e. about $18489.

Marker's note: one mark for the quarterly inputs $r = 0.02$ , $n = 20$ , one for the annuity sum reaching $14578 in (a); one mark for recognising the second phase as lump-sum compound growth (not another annuity), one for the factor $(1.02)^{12}$ , one for the final $18489. Treating the second phase as a continuing annuity, or compounding for the wrong number of quarters, is the trap.

Practise this

Sources & how we know this

Mathematics Advanced Stage 6 Syllabus (2017) — NESA
Mathematics Advanced HSC exam papers — NESA