Recurrence relations and sequences in WACE Mathematics Applications Unit 3

Syllabus dot point

“Use first-order linear recurrence relations to generate sequences and recognise the patterns of growth and decay they produce.”

WACEMathematics ApplicationsUnit 37 min readUpdated 2 Jun 2026

Quick answer

A recurrence relation defines each term from the one before, together with a starting value. The first-order linear form is $t_{n+1} = R t_n + d$ . The multiplier $R$ and the added constant $d$ decide the behaviour: $R = 1$ gives a linear (arithmetic) pattern, $R > 1$ gives growth, and $0 < R < 1$ gives decay towards a limit. Generate the sequence by iterating the rule from the starting term.

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What this dot point is asking
What a recurrence relation is
Generating terms
Finding the equilibrium
Translating a worded situation into a recurrence
Linking to the financial models
Exam-style questions

What this dot point is asking

You must read and use a recurrence, generate terms, and recognise the growth or decay pattern it produces.

What a recurrence relation is

A recurrence relation gives a rule for the next term in terms of the current term, plus an initial value to start. Without the starting term, the rule alone does not pin down the sequence.

First-order linear recurrence

$t_{n+1} = R t_n + d$ , with a given first term (such as $t_0$ or $t_1$ ).

If $d = 0$ and $R = 1$ : constant sequence.
If $R = 1$ and $d \neq 0$ : arithmetic (adds $d$ each step, linear).
If $d = 0$ and $R \neq 1$ : geometric (multiplies by $R$ each step).
If $R \neq 1$ and $d \neq 0$ : mixed (the financial models such as loans and annuities).

This single form underlies the whole finance topic: compound interest ( $d = 0$ ), depreciation, loans ( $d$ negative) and savings ( $d$ positive).

Generating terms

To generate a sequence, substitute the current term into the rule to get the next, repeating from the starting value. This is exactly what a calculator does when you store a value and apply the rule repeatedly.

Read the pattern from R

With $d = 0$ : $R = 1$ is constant, $R > 1$ grows, $0 < R < 1$ decays. When $d \neq 0$ , the multiplier $R$ still controls long-term growth or decay, while $d$ shifts the level. A sequence approaches an equilibrium when $0 < R < 1$ .

Worked example

Iterating a mixed recurrence

A pond holds $5000$ litres. Each week $20\%$ evaporates and then $600$ litres of rain is added. Model the volume and find the first four weekly volumes.

Step 1: Write the recurrence relation

Losing $20\%$ of the current volume each week means $80\%$ remains, so the multiplier is $0.8$ . The rain then adds a fixed $600$ litres, giving the rule $V_{n+1} = 0.8 V_n + 600$ , with $V_0 = 5000$ .

Step 2: Iterate the rule term by term

Substituting each result into the rule produces the next volume.

$V_1 = 0.8(5000) + 600 = 4000 + 600 = 4600$ .

$V_2 = 0.8(4600) + 600 = 3680 + 600 = 4280$ .

$V_3 = 0.8(4280) + 600 = 3424 + 600 = 4024$ .

Step 3: Describe the long-run behaviour

The volume falls each week but by less each time, because the multiplier $0 < R = 0.8 < 1$ drives the sequence towards an equilibrium rather than to zero or infinity. Setting $V = 0.8V + 600$ gives $0.2V = 600$ and $V = 3000$ litres, the steady level the volume approaches in the long run.

Finding the equilibrium

When $0 < R < 1$ , the sequence settles to a long-run value where the next term equals the current term. Set $t = R t + d$ and solve: $t(1 - R) = d$ , so $t = \dfrac{d}{1 - R}$ . This is the same idea as a transition-matrix steady state and as the perpetuity balance in finance.

Translating a worded situation into a recurrence

Most SCSA recurrence questions describe a real process in words, and the skill is turning the words into $t_{n+1} = R t_n + d$ . Identify two things: the multiplier and the regular change. A percentage change gives the multiplier ("falls by $10\%$ " gives $R = 0.9$ , "rises by $5\%$ " gives $R = 1.05$ ); a fixed amount added or removed each period gives $d$ (a $\$300$ deposit gives $d = +300$ , a $\$300$ withdrawal gives $d = -300$ ). Then record the starting value as $t_0$ .

For example, "a tank holds $2000$ litres; each day $5\%$ evaporates and $80$ litres is added" becomes $V_{n+1} = 0.95 V_n + 80$ , with $V_0 = 2000$ . Because $0 < R = 0.95 < 1$ , the volume approaches the equilibrium $V = \dfrac{80}{1 - 0.95} = \dfrac{80}{0.05} = 1600$ litres in the long run.

Linking to the financial models

This single recurrence underlies the whole Unit 3 and Unit 4 finance strand. Compound interest sets $d = 0$ and $R > 1$ ; reducing-balance depreciation sets $d = 0$ and $0 < R < 1$ ; a reducing-balance loan keeps $R > 1$ with $d$ negative (a repayment); a savings annuity keeps $R > 1$ with $d$ positive (a deposit). Recognising which case a question describes tells you immediately how the sequence will behave and which long-run outcome (growth, decay, or equilibrium) to expect.

Common errors

Forgetting the starting term: A recurrence needs both the rule and an initial value; the rule alone is not enough.
Applying operations in the wrong order: Follow the rule exactly: usually multiply by $R$ first, then add $d$ .
Confusing $R$ and $d$ roles: $R$ controls growth or decay; $d$ shifts the level. A multiplier of $1$ is arithmetic; a multiplier away from $1$ is geometric.
Rounding each term: Carry full precision through iteration, especially for financial sequences, and round only at the end.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2021-style5 marks

A sequence is defined by

t_{n+1} = 1.5t_n - 4

, with

t_0 = 20

. (a) Find the first four terms. (b) Describe the long-term behaviour of the sequence and explain why it behaves that way.

Show worked answer →

Apply the rule term by term, then read the multiplier.

(a) $t_0 = 20$ . $t_1 = 1.5(20) - 4 = 30 - 4 = 26$ . $t_2 = 1.5(26) - 4 = 39 - 4 = 35$ . $t_3 = 1.5(35) - 4 = 52.5 - 4 = 48.5$ . So $20, 26, 35, 48.5$ . (3 marks)

(b) The terms increase and grow faster each step. Because the multiplier $1.5 > 1$ , the geometric growth dominates the constant subtraction, so the sequence increases without bound. (2 marks)

Markers reward accurate iteration and linking the growth to the multiplier exceeding $1$ .

WACE 2023-style4 marks

Classify each recurrence as producing a linear, geometric-growth or geometric-decay pattern, giving a reason: (i)

A_{n+1} = A_n + 50

; (ii)

A_{n+1} = 1.08 A_n

; (iii)

A_{n+1} = 0.9 A_n