TASMathematics ApplicationsSyllabus dot point

How do we model a process that both multiplies and adds a fixed amount each step?

Use first-order linear recurrence relations of the form Vn+1 = R Vn + d to model and analyse practical situations.

The general first-order linear recurrence Vn+1 = R Vn + d, its long-term behaviour, equilibrium value, and financial and population applications in TCE Mathematics Applications.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-30

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What this dot point is asking

Arithmetic sequences only add, and geometric sequences only multiply. Many real processes do both at once: a savings account earns interest and also receives a regular deposit; a fish population breeds and is also harvested. The first-order linear recurrence captures both actions in one rule.

The two special cases you already know sit inside this one form. When $R = 1$ the rule becomes $V_{n+1} = V_n + d$ , an arithmetic sequence. When $d = 0$ it becomes $V_{n+1} = R\,V_n$ , a geometric sequence.

Generating terms

You usually build the sequence step by step on a calculator: start at $V_0$ , multiply by $R$ , add $d$ , and repeat. The order matters: multiply first, then add or subtract the constant.

A reducing loan as a recurrence

A loan of $\$ 5000 $charges 1 percent interest per month, and$ \ $300$ is repaid at the end of each month. Track the balance.

Set up and iterate

$R = 1.01$ (interest added), $d = -300$ (repayment removed), $V_0 = 5000$ .

V_{n+1} = 1.01\,V_n - 300

V_1 = 1.01(5000) - 300 = 5050 - 300 = 4750

V_2 = 1.01(4750) - 300 = 4797.50 - 300 = 4497.50

V_3 = 1.01(4497.50) - 300 = 4542.48 - 300 = 4242.48

The balance falls each month because the $\$ 300$ repayment exceeds the interest charged.

Long-term behaviour

What happens after many steps depends on $R$ . If $R = 1$ , the sequence grows or shrinks by a constant $d$ forever (linear). If $R > 1$ , the multiplying effect eventually dominates and the terms grow without bound. If $0 < R < 1$ , the terms settle toward a steady value called the equilibrium or steady state.

Choosing the model

Use a plain geometric model when nothing is added or removed beyond the percentage change. Use the full first-order linear recurrence whenever there is also a fixed regular amount, such as a repayment, deposit, or harvest, layered on top of the percentage change.

A complete answer writes the recurrence in the form $V_{n+1} = R\,V_n + d$ with the correct $R$ and signed $d$ , states $V_0$ , generates the requested terms in the right order, and where asked finds the steady state from $V = d/(1 - R)$ .

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.

2024 TASC General Mathematics9 marksA mining town with a population of 5000 is declining at 15% per year as workers depart, while tourism causes an increase of 600 people per year. a) Find a difference equation to represent the population year by year. b) What happens to the population in the long term? The rate of decline then decreases to 10% (other variables unchanged). c) Find a new difference equation. d) Now what happens in the long term? e) What rate of decline should the council aim for to maintain a steady population of 5000?

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a) (2 marks) Keeping 85% and adding 600: Vn+1 = 0.85 Vn + 600, V0 = 5000.

b) (2 marks) The population falls towards a steady state (equilibrium) where Vn+1 = Vn = V: V = 0.85V + 600, so 0.15V = 600, V = 4000. Long term it settles at 4000 people.

c) (1 mark) With 10% decline R becomes 0.90: Vn+1 = 0.90 Vn + 600, V0 = 5000.

d) (2 marks) Equilibrium: V = 0.90V + 600, so 0.10V = 600, V = 6000. The population now rises towards a steady state of 6000.

e) (2 marks) For a steady 5000, set 5000 = R(5000) + 600, so R(5000) = 4400, R = 0.88. A decline rate of 12% per year keeps the population steady at 5000.

2024 TASC General Mathematics4 marksFollowing are two sequence rules in explicit form. Find the recursive form (difference equation) of each rule. a) Tn = 12 + (n - 1)16. b) Tn = 0.5 x (-3) to the power (n - 1).

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a) (2 marks) This is arithmetic with first term T1 = 12 and common difference d = 16, so each term is the previous term plus 16. Recursive form: Tn+1 = Tn + 16, T1 = 12.

b) (2 marks) This is geometric with first term T1 = 0.5 and common ratio r = -3, so each term is the previous term multiplied by -3. Recursive form: Tn+1 = -3 Tn, T1 = 0.5. Markers want both the rule and the starting term stated.

2024 TASC General Mathematics4 marksA couple has borrowed

400000 at 6.3% p.a. interest compounding monthly. They plan to make monthly payments of

2470. a) Write a difference equation to represent this. Your r value should include a fraction or have at least six decimal places. b) Use your difference equation to determine how long it will take to pay back the loan. Give your answer in months, and comment on the final payment.

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a) (2 marks) Monthly interest factor R = 1 + 0.063/12 = 1.00525. Each month add interest then subtract the payment: Vn+1 = 1.00525 Vn - 2470, V0 = 400000.

b) (2 marks) Iterate the difference equation on the calculator until the balance reaches zero. The balance first becomes negative during the 363rd month, so the loan is repaid in 363 months (about 30 years 3 months). Because the balance overshoots zero, the final payment is smaller than $2470: the amount owing entering month 363 is about$ 1382, which grows to about $1389 with interest, so the last instalment is roughly$ 1389 rather than a full $2470. Markers expect a comment that the final payment is a reduced part payment.