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Year 12 Maths Standard 2 (2027) Annuities

Annuities as Recurrence Relations & Tables

20 practice questions 0 video lessons Theory + worked examples

Master annuities as recurrence relations and tables for NSW Year 12 Mathematics Standard 2. An annuity is a series of equal, regular payments into or out of an account earning compound interest, and you model it period by period with the recurrence \(A_{n+1}=A_n(1+r)+M\) and a table of values of opening balance, interest, payment and closing balance.

You will build the table for up to four periods for a savings annuity (regular deposits) and for a single-sum investment drawn down by regular withdrawals, carry each closing balance forward as the next opening balance, and work out the effect of changing the deposit, rate or duration β€” core annuity skills for Standard 2 financial mathematics.

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Theory

An annuity is a series of equal, regular payments into or out of an account earning compound interest. This Year 12 Standard 2 (NSW) guide shows how to model one with a recurrence relation \(A_{n+1}=A_n(1+r)+M\) and a table of values β€” opening balance, interest, payment and closing balance β€” for savings annuities and for single-sum investments drawn down by regular withdrawals.

An annuity is a sequence of equal, regular payments into (or out of) an account that earns compound interest. A savings annuity has regular deposits; a single-sum investment paying an annuity has regular withdrawals.

You can model an annuity period by period with a recurrence relation. For regular deposits \(M\) at rate \(r\) per period, the balance follows \(A_{n+1}=A_n(1+r)+M\): multiply by \((1+r)\) to add one period's interest, then add the deposit. For regular withdrawals \(W\) you subtract instead: \(A_{n+1}=A_n(1+r)-W\).

In this Year 12 Standard 2 (NSW) topic you set the working out in a table of values β€” period, opening balance, interest, payment and closing balance β€” for up to four periods, and read off the balance or future value. The closing balance of one period is the opening balance of the next.

Balance of a savings annuity growing over timeThe balance rises each year and climbs faster later as interest compounds t A
The balance \(A\) (in \(\$'000\)) of a savings annuity grows each year \(t\), rising faster later as interest compounds.
Effect of a larger regular depositTwo growing balance curves: larger yearly deposits build the balance faster t A $1000/yr $1500/yr
Larger yearly deposits (red, \(\$1500\)) build the balance faster than smaller deposits (navy, \(\$1000\)).

For a balance \(A_n\), interest rate \(r\) per period and regular deposit \(M\), a savings annuity follows the recurrence:

\[A_{n+1} = A_n(1+r) + M\]
An+1=An(1+r)+M

This is the same as adding one period's interest and then the deposit:

\[\text{Interest} = r \times A_{\text{opening}}, \qquad A_{\text{closing}} = A_{\text{opening}} + \text{Interest} + M\]
Interest=r×Aopening

For a single-sum investment with regular withdrawals \(W\), subtract the withdrawal instead:

\[A_{n+1} = A_n(1+r) - W\]
An+1=An(1+r)W
Carry forward. The closing balance of one period is the opening balance of the next, so the balance is tracked step by step down the table of values until you reach the period you need.

How to model an annuity with a table of values

  1. Opening balance. Start period 1 with the initial amount invested (\(A_0\); it is \(0\) if the first payment starts the account).
  2. Interest. Multiply the rate by the opening balance and round to the nearest cent: \(\text{Interest}=r\times A_{\text{opening}}\).
  3. Closing balance. Add the interest, then add the deposit (or subtract the withdrawal): \(A_{\text{closing}}=A_{\text{opening}}+\text{Interest}+M\).
  4. Carry forward and repeat. Use the closing balance as the next opening balance, up to the required period; read the balance or future value from the table.

For example, \(\$1000\) deposited at the end of each year at \(6\%\) p.a.:

YearOpening ($)Interest ($)Deposit ($)Closing ($)
10.000.001000.001000.00
21000.0060.001000.002060.00
Example 1 β€” Build a table of values
\(\$2000\) is deposited at the end of each year into an account earning \(5\%\) p.a. Build a table of values for the first \(4\) years and state the balance after the fourth deposit.
Solution

Use \(A_{n+1}=A_n(1.05)+2000\) with \(A_0=0\): interest \(=0.05\times\) opening; closing \(=\) opening \(+\) interest \(+2000\).

YearOpening ($)Interest ($)Deposit ($)Closing ($)
10.000.002000.002000.00
22000.00100.002000.004100.00
34100.00205.002000.006305.00
46305.00315.252000.008620.25
\(\text{Year 1}\)\(\)\(0+0+2000 = 2000\)
\(\text{Year 2}\)\(\)\(0.05\times2000=100,\ \ 2000+100+2000=4100\)
\(\text{Year 3}\)\(\)\(0.05\times4100=205,\ \ 4100+205+2000=6305\)
\(\text{Year 4}\)\(\)\(0.05\times6305=315.25,\ \ =8620.25\)

The balance after the fourth deposit is \(\$8620.25\).

Example 2 β€” Apply the recurrence
A savings annuity is modelled by \(A_{n+1}=A_n\times1.045+600\) with \(A_0=5000\) (\(\$5000\) already invested at \(4.5\%\) p.a., with \(\$600\) added each year). Find the balance after \(3\) deposits, to the nearest cent.
Solution

Apply the recurrence three times, carrying each closing balance forward.

\(A_1\)\(=\)\(5000\times1.045+600 = 5825\)
\(A_2\)\(=\)\(5825\times1.045+600 = 6687.13\)
\(A_3\)\(=\)\(6687.13\times1.045+600 = 7588.05\)
A3=7588.05

After \(3\) deposits the balance is \(\$7588.05\).

Example 3 β€” Single-sum investment with withdrawals
\(\$50\,000\) is invested at \(4\%\) p.a. and \(\$12\,000\) is withdrawn at the end of each year, so \(A_{n+1}=A_n\times1.04-12000\) with \(A_0=50\,000\). Build a table for \(4\) years and find how much remains after the fourth withdrawal.
Solution

Each year: interest \(=0.04\times\) opening; closing \(=\) opening \(+\) interest \(-12000\).

YearOpening ($)Interest ($)Withdrawal ($)Closing ($)
150000.002000.0012000.0040000.00
240000.001600.0012000.0029600.00
329600.001184.0012000.0018784.00
418784.00751.3612000.007535.36
\(\text{Year 1}\)\(\)\(0.04\times50000=2000,\ \ 50000+2000-12000=40000\)
\(\text{Year 2}\)\(\)\(0.04\times40000=1600,\ \ =29600\)
\(\text{Year 3}\)\(\)\(0.04\times29600=1184,\ \ =18784\)
\(\text{Year 4}\)\(\)\(0.04\times18784=751.36,\ \ =7535.36\)

After the fourth withdrawal \(\$7535.36\) remains.

Example 4 β€” Effect of a larger deposit
In Example 1 (\(\$2000\) each year at \(5\%\) p.a.) the balance after \(4\) years was \(\$8620.25\). If instead \(\$3000\) is deposited each year, find the balance after \(4\) years and how much more interest is earned.
Solution

Rebuild the table with \(A_{n+1}=A_n(1.05)+3000\), \(A_0=0\).

YearOpening ($)Interest ($)Deposit ($)Closing ($)
10.000.003000.003000.00
23000.00150.003000.006150.00
36150.00307.503000.009457.50
49457.50472.883000.0012930.38
\(A_4\)\(=\)\(12930.38\)
\(\text{interest}\)\(=\)\(12930.38-4\times3000 = 930.38\)
\(\text{extra}\)\(=\)\(930.38-620.25 = 310.13\)

The balance is \(\$12930.38\) β€” larger deposits build the balance faster and earn \(\$310.13\) more interest, because interest is earned on a bigger running balance.

Common pitfalls

Interest on the opening balance. Recompute the interest each period on the current opening balance β€” it is not a fixed dollar amount and it is not based on the deposit.
Deposit after interest. Add the regular deposit after the interest; a payment made at the end of a period earns no interest in that same period.
Carry the closing balance forward. The closing balance of one period is the opening balance of the next; do not reset it to the first deposit.

Frequently asked questions

What is an annuity?

An annuity is a sequence of equal, regular payments into or out of an account that earns compound interest. A savings annuity has regular deposits, for example paying the same amount into super each year. A single-sum investment paying an annuity has regular equal withdrawals, for example a retirement fund drawn down each year.

What is the recurrence relation for an annuity?

For regular deposits M at an interest rate r per period the balance follows A(n+1) = A(n) times (1 + r) plus M: you multiply the balance by (1 + r) to add one period's interest, then add the deposit. For regular withdrawals W you subtract instead: A(n+1) = A(n) times (1 + r) minus W.

How do you build an annuity table of values?

Use columns for the period, opening balance, interest, payment and closing balance. Each row: interest equals the rate times the opening balance; closing balance equals opening plus interest plus the deposit (or minus the withdrawal). Carry that closing balance down as the next row's opening balance and repeat for each period, usually up to four.

Do you add the interest before or after the deposit?

Add the interest first, then the deposit. The recurrence A(n+1) = A(n)(1 + r) + M multiplies by (1 + r) before adding M, which matches a deposit paid in at the end of the period. A payment made at the end of a period earns no interest during that same period.

What is the difference between a savings annuity and a single-sum investment with withdrawals?

A savings annuity starts from a small balance and grows because you add equal deposits and earn interest, so you use plus M. A single-sum investment starts with a large lump sum and shrinks because you take equal withdrawals, so you use minus W. Both earn compound interest on the opening balance each period.

Why does the balance grow faster later in a savings annuity?

Interest is earned on the whole opening balance, which gets bigger every period as deposits and past interest build up. So each period earns more interest than the last, and the balance curve rises more steeply the longer the annuity runs. Increasing the regular deposit builds the balance even faster.