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

Annuity investments and superannuation modelled by Vn+1 = Vn(1 + r) + d, with growth of contributions plus interest in TCE Mathematics Applications.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-30

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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.

Iterate

$r = 0.004$ , $d = 500$ , $V_0 = 20000$ .
$V_1 = 20000(1.004) + 500 = 20080 + 500 = 20580$ .
$V_2 = 20580(1.004) + 500 = 20662.32 + 500 = 21162.32$ .
$V_3 = 21162.32(1.004) + 500 = 21246.97 + 500 = 21746.97$ .

After three months the fund holds about $\$ 21,746.97$, growing from both interest and the steady contributions.

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.

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 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.

2023 TASC General Mathematics9 marksA business expects a new vehicle to last about 10 years. To prepare to replace it, the business pays

2500 every quarter into a fund paying 5.8% p.a. compounding quarterly. a) Use a difference equation to model the amount in the fund (use initial term T0). b) How much is in the fund after 10 years? Instead, the company makes an initial deposit of

10000 then the quarterly $2500 instalments. c) Remodel the difference equation. d) How much will be in the account after 10 years? e) What would the present value of the investment's total be?

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a) (2 marks) Quarterly rate i = 0.058/4 = 0.0145. Each quarter the balance earns interest then $2500 is added: Tn+1 = 1.0145 Tn + 2500, T0 = 0.

b) (2 marks) Iterate for 10 years = 40 quarters (or use the future-value annuity formula). The fund reaches about $134244.

c) (2 marks) Only the starting value changes: Tn+1 = 1.0145 Tn + 2500, T0 = 10000.

d) (2 marks) The $10000 also grows for 40 quarters, adding 10000 x 1.0145^40 to the previous total, giving about$ 152030.

e) (1 mark) Present value discounts the final amount back 40 quarters: PV = 152030 / 1.0145^40 = $85477 (approximately). It represents the single deposit today that would grow to the same total.

2024 TASC General Mathematics3 marksImagine you want to save $6000 over 2 years. Your savings account pays 3% p.a. compounding monthly. Use a formula and algebra to find how much you would have to deposit each month.

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

Using the annuities-in-advance future value formula FV = d(1 + i)[(1 + i)^n - 1]/i, substitute FV = 6000:

6000 = d(1.0025)[(1.0025)^24 - 1]/0.0025.

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