First-Order Linear Recurrence Relations - TCE Mathematics Applications (Tasmania)

Syllabus dot point

“Use first-order linear recurrence relations of the form V(n+1) = R V(n) + d to model and analyse practical situations.”

TCEMathematics ApplicationsUnit 38 min readUpdated 2 Jun 2026

Quick answer

You must use the general first-order linear recurrence $V_{n+1} = R\,V_n + d$ , which both multiplies by $R$ and adds $d$ each step, to model situations such as loans and managed populations, generate terms, and reason about long-term behaviour and the steady-state value.

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.

First-order linear recurrence

A rule of the form $V_{n+1} = R\,V_n + d$ , with a given starting value $V_0$ . It is first-order because each term depends only on the one before it, and linear because the previous term is not raised to any power other than one.

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.

The multiplier and the constant

$R$ is the growth or decay factor applied to the running total each step. A rate of $p\%$ per period gives $R = 1 + \tfrac{p}{100}$ for growth or $R = 1 - \tfrac{p}{100}$ for decay. The constant $d$ is the fixed amount added (positive) or removed (negative) after the multiplication.

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 $1\%$ interest per month, and $\$300$ is repaid at the end of each month. We set up the recurrence relation and generate the first three balances.

Step 1: Identify R and d

Each month the outstanding balance is multiplied by the interest factor first, then the repayment is subtracted. Interest at $1\%$ per month gives the multiplier $R = 1.01$ . The repayment reduces the balance, so $d = -300$ . The starting balance is $V_0 = 5000$ :

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

Step 2: Compute the balance after month 1

Apply the recurrence once, multiplying first and then subtracting the repayment:

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

Step 3: Compute the balance after month 2

Use $V_1$ as the new starting value:

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

Step 4: Compute the balance after month 3

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 (which starts at $\$50$ and decreases as the balance falls).

Final answer: After 3 months the outstanding balance is $\$4242.48$ .

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.

Steady-state value

When $0 < R < 1$ , the sequence approaches the equilibrium where $V_{n+1} = V_n = V$ . Solving $V = RV + d$ gives

V = \frac{d}{1 - R}.

At this value the multiply step and the add step exactly cancel, so the total stops changing.

Finding the steady state of a population

A lake holds 8000 fish. Each year the population grows by 20 percent, then 1500 fish are harvested. Find the long-term population.

Model and solve for equilibrium

$R = 1.20$ , $d = -1500$ . Because $R > 1$ here, check the equilibrium $V = \dfrac{d}{1 - R} = \dfrac{-1500}{1 - 1.20} = \dfrac{-1500}{-0.20} = 7500$ .

If the population starts above 7500 it falls toward 7500; if it starts below, it rises toward 7500. Starting at 8000, the population declines gradually toward the steady state of 7500 fish.

Read the sign of d from the words

Deposits, interest credits, and births make $d$ positive; repayments, withdrawals, and harvests make $d$ negative. Decide the sign before iterating, because one wrong sign sends the whole sequence the wrong way.

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.

Adding the constant before multiplying

The rule is multiply by $R$ first, then apply $d$ . Doing it the other way, $R(V_n + d)$ , charges interest on the repayment or grows the deposit by a year it was never invested. Always compute $R\,V_n$ first, then add $d$ .

Converting between recursive and explicit forms

TASC questions often give a rule in explicit (closed) form and ask for the recursive form, or the reverse. The link is direct. An explicit arithmetic rule $t_n = a + (n-1)d$ comes from the recurrence $t_{n+1} = t_n + d$ with $t_1 = a$ . An explicit geometric rule $t_n = a r^{\,n-1}$ comes from $t_{n+1} = r\,t_n$ with $t_1 = a$ . To go from explicit to recursive, read off the multiplier (the base of the power, which becomes $R$ ) and the constant added each step (which becomes $d$ ), then state the starting term. To go the other way, identify $R$ and $d$ and recognise the matching closed form, or simply iterate.

Recursive to explicit

A savings balance follows $V_{n+1} = 1.05\,V_n$ with $V_0 = 2000$ (no deposits).

Recognise the form

With $d = 0$ and $R = 1.05$ this is purely geometric, so the explicit form is $V_n = V_0 R^n = 2000(1.05)^n$ .

Check

$V_1 = 2000(1.05) = 2100$ , matching one step of the recurrence. The closed form lets you jump straight to, say, $V_{10} = 2000(1.05)^{10} = \$3257.79$ without listing every year.

Reading the steady state from a graph

When a first-order linear recurrence with $0 < R < 1$ is plotted, the points approach a horizontal line at the equilibrium value $V = \dfrac{d}{1 - R}$ . If the start is above this line the sequence falls toward it; if below, it rises toward it. The equilibrium does not depend on the starting value, only on $R$ and $d$ , which is why two funds with the same rate and the same regular amount end up at the same long-run level no matter where they begin.

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, converts between recursive and explicit forms when asked, and finds the steady state from $V = d/(1 - R)$ .

Exam-style questions

Questions in the style of TASC 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.

Practice9 marks

A mining town of 5000 is declining at

15\%

per year as workers depart, while tourism adds 600 people per year. a) Find a difference equation for the population. b) What happens in the long term? The decline then eases to

10\%

. c) Find the new difference equation. d) Now what happens long term? e) What decline rate keeps the population steady at 5000?

Show worked answer →

a) (2 marks) Keeping $85\%$ and adding 600: $V_{n+1} = 0.85\,V_n + 600$ , $V_0 = 5000$ .

b) (2 marks) The population settles at the equilibrium where $V_{n+1} = V_n = V$ : $V = 0.85V + 600$ , so $0.15V = 600$ , $V = 4000$ . Long term it settles at 4000 people.

c) (1 mark) With $10\%$ decline $R$ becomes $0.90$ : $V_{n+1} = 0.90\,V_n + 600$ , $V_0 = 5000$ .

d) (2 marks) Equilibrium: $V = 0.90V + 600$ , so $0.10V = 600$ , $V = 6000$ . The population now rises towards 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 holds the population at 5000.

Practice4 marks

Find the recursive form (difference equation) of each rule. a)

t_n = 12 + (n - 1)16

. b)

t_n = 0.5 \times (-3)^{n-1}

Show worked answer →

a) (2 marks) This is arithmetic with first term $t_1 = 12$ and common difference $d = 16$ , so each term is the previous term plus 16. Recursive form: $t_{n+1} = t_n + 16$ , $t_1 = 12$ .

b) (2 marks) This is geometric with first term $t_1 = 0.5$ and common ratio $r = -3$ , so each term is the previous term times $-3$ . Recursive form: $t_{n+1} = -3\,t_n$ , $t_1 = 0.5$ . Markers want both the rule and the starting term stated.

Practice4 marks

A couple borrows

\$400\,000

6.3\%

p.a. compounding monthly, with monthly payments of

\$2470

. a) Write a difference equation. Your

R

value should be a fraction or have at least six decimal places. b) How long to repay the loan? Give your answer in months and comment on the final payment.

Show worked answer →

a) (2 marks) Monthly interest factor $R = 1 + \frac{0.063}{12} = 1.00525$ . Each month add interest then subtract the payment: $V_{n+1} = 1.00525\,V_n - 2470$ , $V_0 = 400000$ .

b) (2 marks) Iterate 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 it 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 $\$1389$ . Markers expect a comment that the final payment is a reduced part payment.

Practise this

Sources & how we know this

General Mathematics Level 3 (MTG315123) course document (2023 onwards) — TASC