Sequences and recurrence relations: VCE General Mathematics Unit 3 Recursion and financial modelling

Syllabus dot point

“Use a first-order linear recurrence relation of the form u(0) = a, u(n+1) = R u(n) + d to generate the terms of a sequence, recognise arithmetic and geometric sequences, use the explicit rules, and describe the behaviour of the sequence numerically and graphically”

15 min readUpdated 23 Sep 2026

Quick answer

The recurrence $u_0 = a$ , $u_{n+1} = Ru_n + d$ generates every sequence in the recursion topic. With $R = 1$ it is arithmetic, $u_n = u_0 + nd$ (simple interest, flat rate and unit cost depreciation); with $d = 0$ it is geometric, $u_n = R^nu_0$ (compound interest, reducing balance depreciation); otherwise it is the mixed model for loans and annuities. Count from $u_0$ , show each recursive step when asked, and describe behaviour as increasing, decreasing, oscillating or levelling off.

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What this dot point is asking
The answer
Exam-style questions
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What this dot point is asking

Every model in Recursion and financial modelling (depreciation, simple and compound interest, loans, annuities, perpetuities and annuity investments) is built on one formula, printed on the formula sheet:

u_0 = a, \qquad u_{n+1} = Ru_n + d.

This is a first-order linear recurrence relation: each term is found from the one before it by multiplying by a constant $R$ and adding a constant $d$ , starting from the initial value $u_0 = a$ . VCAA wants you to generate terms from it (by hand, with "recursive calculations" shown, and on the CAS), to recognise the arithmetic case ( $R = 1$ ) and the geometric case ( $d = 0$ ), to use their explicit rules, and to describe how a sequence behaves (increasing, decreasing, oscillating, levelling off) from a table or graph.

The answer

Generating terms

Start from $u_0$ and apply the rule repeatedly:

u_1 = Ru_0 + d, \quad u_2 = Ru_1 + d, \quad u_3 = Ru_2 + d, \ldots

For $u_0 = 100$ and $u_{n+1} = 0.5u_n + 20$ : $u_1 = 0.5 \times 100 + 20 = 70$ , $u_2 = 0.5 \times 70 + 20 = 55$ , $u_3 = 0.5 \times 55 + 20 = 47.5$ .

Note the numbering: in VCE the first term is $u_0$ , so $u_2$ is the value after two applications of the rule. The 2023 question on $T_2$ was answered correctly by only $43\%$ of students, largely because of this.

On a CAS you can generate terms quickly by typing the starting value, then repeatedly entering "previous answer $\times R + d$ ", or by using a sequence or spreadsheet feature. In a "show that" or "showing recursive calculations" question you must still write each step.

Showing recursive calculations

When a question says "showing recursive calculations", the examiners want each application of the rule written with the numbers:

A_1 = 1.016 \times 30\,000 - 1515.18 = 28\,964.82

A_2 = 1.016 \times 28\,964.82 - 1515.18 = 27\,913.08

The 2023 report is explicit: the mark "could only be obtained by showing the correct recursive, step-by-step approach", and students also lost the mark by rounding the intermediate value too early. Keep money values to the cent at each step.

The arithmetic case: $R = 1$

If $R = 1$ , the rule is $u_{n+1} = u_n + d$ : the same amount is added (or subtracted, if $d < 0$ ) each time. The sequence is arithmetic, and its terms lie on a straight line when graphed. After $n$ steps, $n$ lots of $d$ have been added:

u_n = u_0 + nd.

Arithmetic sequences model simple interest, flat rate depreciation and unit cost depreciation. For $u_0 = 7$ , $u_{n+1} = u_n + 4$ : $u_n = 7 + 4n$ , so $u_{10} = 47$ .

The geometric case: $d = 0$

If $d = 0$ , the rule is $u_{n+1} = Ru_n$ : each term is multiplied by the same factor. The sequence is geometric, and its terms lie on a curve. After $n$ steps the starting value has been multiplied by $R$ a total of $n$ times:

u_n = R^n u_0.

Geometric sequences model compound interest ( $R > 1$ ), reducing balance depreciation ( $0 < R < 1$ ) and quantities like the height of a bouncing ball. For $u_0 = 2$ , $u_{n+1} = 3u_n$ : $u_n = 3^n \times 2$ , so $u_5 = 486$ .

Key fact

Formula sheet: $u_0 = a$ , $u_{n+1} = Ru_n + d$ . If $R = 1$ the sequence is arithmetic, $u_n = u_0 + nd$ (linear growth or decay). If $d = 0$ it is geometric, $u_n = R^nu_0$ (exponential growth if $R > 1$ , decay if $0 < R < 1$ , oscillation if $R < 0$ ). If $R \ne 1$ and $d \ne 0$ it is neither: this mixed case models loans, annuities and annuity investments. The first term is $u_0$ , so the powers and multiples are $n$ , not $n - 1$ .

The mixed case: $R \ne 1$ and $d \ne 0$

When both $R \ne 1$ and $d \ne 0$ , the sequence is neither arithmetic nor geometric. This is the model for:

reducing balance loans and annuities: $R > 1$ , $d < 0$ (interest added, payment subtracted);
annuity investments: $R > 1$ , $d > 0$ (interest added, deposit added);
perpetuities: $R > 1$ with $d$ exactly cancelling the interest, so the sequence is constant.

For these, generate terms with the recurrence (or use the finance solver); the study design does not require an explicit rule.

Describing the behaviour

Look at a table or graph of the terms and describe what happens as $n$ increases:

Behaviour	Typical recurrence	Example
Increasing linearly	$R = 1$ , $d > 0$	$u_{n+1} = u_n + 4$
Decreasing linearly	$R = 1$ , $d < 0$	$u_{n+1} = u_n - 25$
Increasing exponentially	$R > 1$ , $d = 0$ , $u_0 > 0$	$u_{n+1} = 1.1u_n$
Decaying towards 0	$0 < R < 1$ , $d = 0$	$u_{n+1} = 0.5u_n$
Oscillating	$R < 0$	$u_{n+1} = -u_n$
Levelling off at a limit	$0 < R < 1$ , $d \ne 0$	$u_{n+1} = 0.5u_n + 20$
Constant	$u_{n+1} = u_n$ (for example a perpetuity)	$u_{n+1} = 1.01u_n - 1000$ , $u_0 = 100\,000$

For $u_{n+1} = 0.5u_n + 20$ , the terms approach 40 whatever the starting value. At the limit the terms stop changing, so $L = 0.5L + 20$ , giving $L = 40$ . (Limiting values are a Units 1 and 2 idea that remains useful for interpreting Unit 3 models, such as the long-run value of an annuity or a population.)

Recognising the type from terms

Given some terms, test for a constant difference (arithmetic) or a constant ratio (geometric). If neither is constant, try to find $R$ and $d$ from three consecutive terms: $u_1 = Ru_0 + d$ and $u_2 = Ru_1 + d$ give two equations. For 20, 34, 55: subtracting gives $21 = 14R$ , so $R = 1.5$ , and $d = 34 - 30 = 4$ .

Worked examples: generating, explicit rules and behaviour

Oscillating sequence

$T_0 = 5$ , $T_{n+1} = -T_n$ . Find $T_2$ and describe the sequence.

Generate. $T_1 = -5$ , $T_2 = 5$ .

Describe. Geometric with $R = -1$ : it oscillates between 5 and $-5$ .

Marker's note: count from $T_0$ . $T_2$ is the result of two steps.

Explicit rules

(a) $u_0 = 7$ , $u_{n+1} = u_n + 4$ . (b) $u_0 = 2$ , $u_{n+1} = 3u_n$ . Find $u_{10}$ and $u_5$ respectively.

(a) Arithmetic: $u_n = 7 + 4n$ , $u_{10} = 47$ .

(b) Geometric: $u_n = 3^n \times 2$ , $u_5 = 486$ .

Marker's note: substitute $n = 0$ to check: each rule must return the starting value.

A bouncing ball

A ball dropped from 2 m rebounds to $75\%$ of its previous height. When is the rebound height first below 0.5 m?

Model: $h_0 = 2$ , $h_{n+1} = 0.75h_n$ , so $h_n = 0.75^n \times 2$ .
Table: $h_4 \approx 0.633$ , $h_5 \approx 0.475$ .
Answer: After the 5th bounce.

Marker's note: a CAS table of $0.75^n \times 2$ is the quickest method.

Recursive calculations for a loan

$A_0 = 30\,000$ , $A_{n+1} = 1.016A_n - 1515.18$ . Find $A_2$ .

$A_1 = 1.016 \times 30\,000 - 1515.18 = 28\,964.82$ .

$A_2 = 1.016 \times 28\,964.82 - 1515.18 = 27\,913.08$ .

Marker's note: both lines are needed for the mark, and $A_1$ must be kept to the cent.

Common traps

Starting from $u_1$: VCE sequences start at $u_0$ . The explicit rules are $u_n = u_0 + nd$ and $u_n = R^nu_0$ , with $n$ , not $n - 1$ .
Calling the mixed case geometric: $u_{n+1} = 1.05u_n - 500$ is neither arithmetic nor geometric.
Mixing up the depreciation types: Flat rate and unit cost are arithmetic; reducing balance is geometric.
Skipping steps: "Showing recursive calculations" requires every step written out.
Rounding intermediate values: Round only the final answer (or to the cent in money steps).
Misreading $R$: A decrease of $12\%$ per step is $R = 0.88$ , not $R = 0.12$ or $R = -0.12$ .

Exam technique

Before calculating, classify the recurrence: $R = 1$ (arithmetic), $d = 0$ (geometric) or mixed. In Examination 1, many recurrence questions are answered by generating two or three terms on the CAS and matching them to the options. In Examination 2, write every recursive step when asked to show it, and give money to the cent at each step.

Note

A recurrence relation is a recipe: start with a number, then keep doing the same thing to get the next one, like "multiply by 1.05 and take away 500". If the recipe only adds or subtracts the same amount, the numbers go up or down in a straight line. If it only multiplies, they grow or shrink faster and faster, like interest or a bouncing ball. If it does both, you get the pattern of a loan being paid off. The first number is called $u_0$ , so after two steps you are at $u_2$ .

Exam-style questions

Questions in the style of VCAA 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.

2023 VCAA-style1 mark

A sequence of numbers is generated by the recurrence relation

T_0 = 5

T_{n+1} = -T_n

. The value of

T_2

is A. -10 B. -5 C. 0 D. 5 E. 10

Show worked answer →

Apply the rule one step at a time from $T_0 = 5$ : $T_1 = -T_0 = -5$ and $T_2 = -T_1 = 5$ . The answer is D.

The rule multiplies by $R = -1$ , so the sequence oscillates: 5, $-5$ , 5, $-5$ , ... Only $43\%$ of students chose D; a common error was to start counting at $T_1$ instead of $T_0$ .

Source: VCAA 2023 General Mathematics Examination 1, Question 17, and the 2023 examination report.

2025 VCAA-style1 mark

A recurrence relation is of the form

u_0 = a

u_{n+1} = Ru_n + d

. If

a > 0

R = 0.5

and

d = 0

, the sequence generated will be A. arithmetic and increasing. B. arithmetic and decreasing. C. geometric and increasing. D. geometric and decreasing.

Show worked answer →

With $d = 0$ the rule is $u_{n+1} = 0.5u_n$ : each term is a constant multiple of the previous one, so the sequence is geometric. Since $a > 0$ and $0 < R < 1$ , each term is half the one before, so it is decreasing. The answer is D ( $53\%$ correct).

Source: VCAA 2025 General Mathematics Examination 1, Question 18, and the 2025 examination report.

2023 VCAA-style1 mark

Arthur borrowed $30 000 for a motorcycle, with interest charged at 6.4% per annum compounding quarterly. The balance after

n

quarters is modelled by

A_0 = 30\,000

A_{n+1} = 1.016A_n - 1515.18

. Showing recursive calculations, determine the balance of the loan after two quarters, rounded to the nearest cent.

Show worked answer →

$A_1 = 1.016 \times 30\,000 - 1515.18 = 30\,480 - 1515.18 = 28\,964.82$

$A_2 = 1.016 \times 28\,964.82 - 1515.18 = 29\,428.257\ldots - 1515.18 = 27\,913.08$

The balance after two quarters is $27 913.08.

The examiners reported that many students lost this mark because they did not show the recursive calculations, rounded too early (28 964.8 instead of 28 964.82), or made transcription errors. "An answer on its own was not sufficient."

Source: VCAA 2023 General Mathematics Examination 2, Question 5b, and the 2023 examination report.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A sequence is generated by

u_0 = 7

u_{n+1} = u_n + 4

. (a) Write the first four terms. (b) Write the explicit rule for

u_n

and find

u_{10}

Show worked solution →

(a) $u_0 = 7$ , $u_1 = 11$ , $u_2 = 15$ , $u_3 = 19$ . (1 mark)

(b) Arithmetic with common difference 4, starting from $u_0$ : $u_n = 7 + 4n$ , so $u_{10} = 7 + 40 = 47$ . (1 mark)

foundation2 marks

A sequence is generated by

u_0 = 2

u_{n+1} = 3u_n

. (a) Name the type of sequence. (b) Find

u_5

using the explicit rule.

Show worked solution →

(a) Geometric (each term is multiplied by 3), increasing. (1 mark)

(b) $u_n = 3^n \times 2$ , so $u_5 = 243 \times 2 = 486$ . (1 mark)

foundation2 marks

For each recurrence relation, state whether the sequence is arithmetic, geometric or neither: (a)

u_{n+1} = u_n - 12

(b)

u_{n+1} = 1.05u_n

(c)

u_{n+1} = 1.05u_n - 500

(d)

u_{n+1} = 0.8u_n

Show worked solution →

(a) Arithmetic ( $R = 1$ , $d = -12$ ). (b) Geometric ( $d = 0$ , $R = 1.05$ ). (c) Neither: both $R \ne 1$ and $d \ne 0$ (this is the loan or annuity model). (d) Geometric ( $R = 0.8$ ). (2 marks: all four correct; 1 mark for three.)

core3 marks

A ball is dropped from 2 m and each bounce reaches 75% of the previous height. (a) Write a recurrence relation for

h_n

, the height after

n

bounces. (b) Find the height after 4 bounces, correct to three decimal places. (c) After how many bounces is the height first below 0.5 m?

Show worked solution →

(a) $h_0 = 2$ , $h_{n+1} = 0.75h_n$ . (1 mark)

(b) $h_4 = 0.75^4 \times 2 = 0.3164\ldots \times 2 \approx 0.633$ m. (1 mark)

(c) $h_4 \approx 0.633$ , $h_5 \approx 0.475$ . First below 0.5 m after 5 bounces. (1 mark)

core3 marks

A sequence is defined by

u_0 = 100

u_{n+1} = 0.5u_n + 20

. (a) Find

u_1

u_2

and

u_3

. (b) Describe the long-term behaviour. (c) Find the value the sequence approaches.

Show worked solution →

(a) $u_1 = 50 + 20 = 70$ , $u_2 = 35 + 20 = 55$ , $u_3 = 27.5 + 20 = 47.5$ . (1 mark)

(b) The terms decrease, by smaller and smaller amounts, and level off: the sequence approaches a limiting value. (1 mark)

(c) In the long run $u_{n+1} \approx u_n = L$ , so $L = 0.5L + 20$ , giving $L = 40$ . (1 mark)

core2 marks

A loan is modelled by

A_0 = 30\,000

A_{n+1} = 1.016A_n - 1515.18

. Showing recursive calculations, find

A_1

and

A_2

to the nearest cent.

Show worked solution →

$A_1 = 1.016 \times 30\,000 - 1515.18 = 28\,964.82$ . (1 mark)

$A_2 = 1.016 \times 28\,964.82 - 1515.18 \approx 27\,913.08$ . (1 mark) Write both lines; do not round $A_1$ to fewer than two decimal places before using it.

exam3 marks

A recurrence relation has the form

u_0 = a

u_{n+1} = Ru_n + d

with

a = 400

. For each case, describe the sequence generated: (a)

R = 1

d = -25

(b)

R = 1.1

d = 0

(c)

R = -1

d = 0

Show worked solution →

(a) Arithmetic and decreasing: 400, 375, 350, ..., falling by 25 each step (a straight-line pattern, like flat rate depreciation). (1 mark)

(b) Geometric and increasing: 400, 440, 484, ..., growing by 10% each step (like compound interest). (1 mark)

(c) Geometric and oscillating: 400, $-400$ , 400, $-400$ , ..., alternating in sign. (1 mark)

exam2 marks

The first three terms of a sequence generated by

u_{n+1} = Ru_n + d

are

u_0 = 20

u_1 = 34

u_2 = 55

. Find

R

and

d

Show worked solution →

$34 = 20R + d$ and $55 = 34R + d$ . Subtracting: $21 = 14R$ , so $R = 1.5$ . (1 mark)

Then $d = 34 - 20 \times 1.5 = 34 - 30 = 4$ . (1 mark) Check: $1.5 \times 34 + 4 = 55$ .

Sequences and recurrence relations: VCE General Mathematics Unit 3 Recursion and financial modelling

What this dot point is asking

The answer

Generating terms

Showing recursive calculations

The arithmetic case: $R = 1$

The geometric case: $d = 0$

The mixed case: $R \ne 1$ and $d \ne 0$

Describing the behaviour

Recognising the type from terms

Oscillating sequence

Explicit rules

A bouncing ball

Recursive calculations for a loan

Exam-style questions

Practice questions

Practise this

Sources & how we know this

What this dot point is asking

The answer

Generating terms

Showing recursive calculations

The arithmetic case: R=1R = 1R=1

The geometric case: d=0d = 0d=0

The mixed case: R≠1R \ne 1R=1 and d≠0d \ne 0d=0

Describing the behaviour

Recognising the type from terms

Oscillating sequence

Explicit rules

A bouncing ball

Recursive calculations for a loan

Exam-style questions

Practice questions

Practise this

Sources & how we know this

The arithmetic case: $R = 1$

The geometric case: $d = 0$

The mixed case: $R \ne 1$ and $d \ne 0$