What is making the deposit the subject?

A common task gives a savings target and asks for the required regular deposit. Rearrange the future-value formula to make d the subject by dividing the target FV by the annuity factor ((1 + i)^n - 1)/(i) (times (1 + i) for an annuity in advance). This is exactly the algebra in the second exam answer above, and the marker rewards showing the rearrangement, not just the final number.

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TASMathematics ApplicationsSyllabus dot point

How does a fund grow when we both earn interest and add regular contributions?

Model annuity investments and superannuation where regular contributions are added to a fund that earns compound interest.

Annuity investments and superannuation modelled by a recurrence that adds interest then a regular contribution, with growth of deposits plus interest in TCE Mathematics Applications.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

An annuity investment is the savings mirror of a loan. Rather than paying a debt down, you build a balance up: the fund earns interest and you add a regular deposit on top. Superannuation is the most common example, where contributions accumulate over a working life.

The recurrence relation

The order of operations is interest first, then the contribution. The plus sign on $d$ is what separates a growing investment from a draining annuity.

Building a superannuation balance

A fund starts at $\$20\,000$ , earns 0.4 percent per month, and receives $\$500$ at the end of each month.

Step 1: Identify the parameters and set up the recurrence

The monthly interest rate is $r = 0.004$ , the regular contribution is $d = 500$ , and the starting balance is $V_0 = 20000$ . Each month we apply interest to the existing balance first and then add the contribution, which matches the recurrence $V_{n+1} = V_n(1.004) + 500$ . Setting up these values before calculating prevents errors later.

Step 2: Iterate the recurrence across three months

Apply the rule three times in sequence, carrying forward the exact result each time.
$V_1 = 20000(1.004) + 500 = 20080 + 500 = 20580$ .
$V_2 = 20580(1.004) + 500 = 20662.32 + 500 = 21162.32$ .
$V_3 = 21162.32(1.004) + 500 = 21246.97 + 500 = 21746.97$ .

Step 3: Interpret the result

After three months the fund holds about $\$21\,746.97$ , growing from both interest and the steady contributions. Notice that each month's interest is calculated on a slightly larger balance than the month before, so the interest portion itself increases over time even though the contribution stays constant.

Interest versus contributions

Early on, most of the growth comes from the contributions, because the balance is small and earns little interest. Over many years the balance becomes large enough that the interest it earns can exceed the contributions, which is the power of compounding over a long working life.

Future value over many periods

To find the balance after a long time, iterate the recurrence with technology or a financial solver. The future value depends on the starting balance, the rate, the contribution, and the number of periods. Increasing any of these increases the final balance, but the rate and the number of periods have the most dramatic effect because they drive compounding.

The future-value formula

For many periods, iterating by hand is slow, so an explicit future-value formula is used. For an ordinary annuity (contributions at the end of each period),

FV = d\,\frac{(1 + i)^n - 1}{i},

where

d

is the regular contribution,

i

is the per-period rate, and

n

is the number of contributions. If the contributions are made at the start of each period (an annuity in advance), multiply the whole expression by an extra

(1 + i)

, because every deposit then earns one more period of interest. Any starting balance

V_0

grows separately as

V_0(1 + i)^n

and is added on.

Future value of a savings plan

Deposit $\$300$ at the end of each month into an account earning $4.8\%$ p.a. compounding monthly for 5 years.

Step 1: Identify and convert the inputs

$d = 300$ , $i = \frac{0.048}{12} = 0.004$ , $n = 5 \times 12 = 60$ . The annual rate must be divided by 12 to match the monthly compounding period, and the time in years must be converted to the same number of monthly periods. Mismatching the rate and period units is the most common error in annuity calculations.

Step 2: Apply the future-value formula and evaluate

Because deposits occur at the end of each month, this is an ordinary annuity and the formula applies directly without adjustment. Substitute the identified values and simplify step by step.

FV = 300 \times \frac{(1.004)^{60} - 1}{0.004} = 300 \times \frac{1.27024 - 1}{0.004} = 300 \times 67.56 = \$20\,268.

Step 3: Interpret the result

After 5 years the plan is worth about $\$20\,268$ , of which $\$18\,000$ is your own deposits and the rest is interest. This demonstrates how compounding amplifies even modest regular deposits over a multi-year period, turning a straightforward saving habit into meaningfully more wealth than the sum of contributions alone.

Making the deposit the subject

A common task gives a savings target and asks for the required regular deposit. Rearrange the future-value formula to make $d$ the subject by dividing the target $FV$ by the annuity factor $\frac{(1 + i)^n - 1}{i}$ (times $(1 + i)$ for an annuity in advance). This is exactly the algebra in the second exam answer above, and the marker rewards showing the rearrangement, not just the final number.

A complete answer models the fund with $V_{n+1} = V_n(1 + r) + d$ using a positive contribution, applies interest before adding the contribution, and finds the future value by iterating or by the future-value formula over the correct number of periods.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20239 marksA business pays

\

2500

every quarter into a fund paying

5.8\%

p.a. compounding quarterly. a) Model the fund with a difference equation (use

T_0

). b) How much is in the fund after 10 years? The company instead makes an initial deposit of

10\,000

then the quarterly instalments. c) Remodel. d) How much after 10 years? e) What is the present value of the total?

Show worked answer →

a) (2 marks) Quarterly rate $i = \frac{0.058}{4} = 0.0145$ . Each quarter the balance earns interest then $\$2500$ is added: $T_{n+1} = 1.0145\,T_n + 2500$ , $T_0 = 0$ .

b) (2 marks) Iterate for 10 years $= 40$ quarters (or use the future-value annuity formula). The fund reaches about $\$134\,244$ .

c) (2 marks) Only the starting value changes: $T_{n+1} = 1.0145\,T_n + 2500$ , $T_0 = 10000$ .

d) (2 marks) The $\$10\,000$ also grows for 40 quarters, adding $10000 \times 1.0145^{40}$ to the previous total, giving about $\$152\,030$ .

e) (1 mark) Present value discounts the final amount back 40 quarters: $PV = \frac{152030}{1.0145^{40}} = \$85\,477$ (approximately). It is the single deposit today that would grow to the same total.

TCE 20243 marksYou want to save

\

6000

over 2 years. Your account pays

3\%$ p.a. compounding monthly. Use a formula and algebra to find how much you must deposit each month.

Show worked answer →

(3 marks) This is a future-value annuity (in advance). Monthly rate $i = \frac{0.03}{12} = 0.0025$ and the number of deposits $n = 24$ .

Using the annuities-in-advance future-value formula $FV = d(1 + i)\frac{(1 + i)^n - 1}{i}$ , substitute $FV = 6000$ :

$6000 = d(1.0025)\frac{(1.0025)^{24} - 1}{0.0025}$ .

Compute the bracket: $(1.0025)^{24} - 1 = 0.06176$ , so the factor is $1.0025 \times \frac{0.06176}{0.0025} = 24.7646$ . Then $d = \frac{6000}{24.7646} = \$242.28$ per month. Markers want the substitution, the algebra to make $d$ the subject, and the rounded monthly deposit.