Topic guide 5 · Unit 3 Topic 4 · Growth and decay in sequences
An arithmetic sequence goes up or down by the same amount every step: a tank losing 135 L a minute, a tractor losing $2700 of value a year, a theatre adding 8 seats to each row. That fixed step is the common difference, $d$. One formula, $t_n = t_1 + (n-1)d$, handles almost every question. The marks go missing in the setup, in deciding which value is $t_1$ and which term number you want.
This guide shows you how to spot an arithmetic sequence, build one from a recursive rule, use the $n$th term rule forwards and backwards, and place $t_1$ correctly in a real context. All twelve arithmetic sequence questions from the 2020 to 2025 externals are worked here, mark by mark from the marking guides. Tick off each skill once you can do it without help.
Part 1
Growth and decay in sequences, arithmetic and geometric together, carried 65 short-response marks from 2020 to 2025 in questions set mainly on sequences, more than any other topic. Count 2023 Paper 2 Question 5, where a sequence supplies the road lengths in a network question, and it is 72. The topic was examined in every one of the six years.
For arithmetic sequences alone I count twelve questions worth 35 marks: five multiple choice, five Paper 1 short responses and two Paper 2 questions. There was a multiple choice question in five of the six years (none in 2024), a Paper 1 short response in five of the six (none in 2023), and a Paper 2 question that uses an arithmetic sequence in every year.
Three more Paper 2 questions pair an arithmetic model with a geometric one. Their harder half is the geometric model, so they are worked in full in the geometric sequences guide. Part 6 shows where the arithmetic half fits.
Unit 3 Topic 4, Growth and decay in sequences, gives the arithmetic sequence 5 hours and four dot points:
The last dot point names three contexts: simple interest, a taxi fare made of a flag fall plus a charge per kilometre, and straight-line depreciation. Depreciation was the whole of 2021 Paper 1 Question 20.
In the 2019 syllabus, which most of these past papers were set under, the sequences topic was Topic 3 and the arithmetic sequence had 4 hours. Its depreciation example also named the unit cost method, which the 2025 syllabus no longer lists. The skills and the $n$th term rule are the same, so every past question on this page still fits.
| Section | Supplied |
|---|---|
| Sequences | arithmetic sequence $t_n = t_1 + (n-1)d$ |
| Sequences | geometric sequence $t_n = t_1 r^{(n-1)}$ |
The book prints the formula only. The meanings come from the syllabus: $t_n$ is the $n$th term, $t_1$ is the first term, $n$ is the term number and $d$ is the common difference. The formula sheet used for the 2020 to 2024 papers printed the same arithmetic rule.
| Question | What it tests |
|---|---|
| 2020 P1 Q3 (MC) 1 mark | Common difference of $4, 2, 0, -2, -4$. Answer $-2$. |
| 2025 P1 Q2 (MC) 1 mark | Three consecutive arithmetic terms. 76.5% correct. |
| 2025 P1 Q21 4 marks | Gym customers. Justify arithmetic ($d = 12$), then plot six days. |
| 2022 P1 Q4 (MC) 1 mark | Graph of $T_{n+1} = T_n + 2$, $T_1 = 4$. 63.6% correct. |
| 2024 P1 Q16 3 marks | Theatre rows from a recursive rule. Complete a table, then total it (148). |
| 2023 P1 Q9 (MC) 1 mark | 6th term of 3, 9, … Answer 33. 75.4% correct. |
| 2020 P1 Q18 4 marks | Exhibition attendees. $d$ from the 1st and 3rd terms (43), then the 6th term (568). |
| 2022 P1 Q18 4 marks | Total songs on a playlist, read from a graph. Predict week 25 (3015). |
| 2021 P1 Q9 (MC) 1 mark | Staircase rule. Units and where $t_1$ sits. 17.5% correct. |
| 2021 P1 Q20 4 marks | Tractor, straight-line depreciation. $20 800. |
| 2020 P2 Q1 5 marks | Tank draining to 5000 L. Solve for $n$, about 56 minutes. |
| 2025 P2 Q6 6 marks | Two hikers on two arithmetic models. Speed 2.4 km/h. |
| 2021 P2 Q4 6 marks | A linear model growing by $126 a year against a compound model read from a graph. |
| 2022 P2 Q6 7 marks | A pattern of lines: slopes geometric, $y$-intercepts arithmetic. Find an intersection. |
| 2023 P2 Q5 7 marks | Road lengths from $t_n = t_1 + 2(n-1)$ inside a Hamiltonian cycle and timing problem. |
| 2024 P2 Q6 6 marks | Meals arithmetic, accommodation geometric. A 7-day cost against an estimate. |
Select a question to jump to its worked solution. The last four are worked in guide 6 or guide 11. Paper 1 is all simple familiar. QCAA does not publish the band of these Paper 2 questions. Percentages are the share of the state that chose the key, from the subject reports (none were published for 2020).
Part 2
An arithmetic sequence adds the same number each time. That number is the common difference, any term minus the one before it:
$$d = t_{n+1} - t_n$$
$d$ can be negative, for a falling sequence, and it can be a decimal. A geometric sequence multiplies by the same number each time instead. The 2020 report listed this difference as something to strengthen, and in 2025 QCAA gave it two marks of its own.
| Arithmetic | Geometric | |
|---|---|---|
| Test | subtract: $t_2 - t_1$, $t_3 - t_2$, … | divide: $t_2 \div t_1$, $t_3 \div t_2$, … |
| What stays the same | common difference $d$ | common ratio $r$ |
| Growth | linear, the same amount each step | the same percentage each step |
| Graph of the terms | dots on a straight line | dots on a curve |
| Example | 6, 18, 30, 42 ($d = 12$) | 3.5, 7.0, 14.0 ($r = 2$) |
$d$ is the later term minus the earlier one. For 4, 2, 0, … that is $2 - 4 = -2$. A falling quantity (depreciation, draining, cooling) always has a negative $d$. A positive $d$ next to a falling quantity means you subtracted backwards.
For the sequence 4, 2, 0, –2, –4 … the common difference is
Subtract each term from the one after it:
$$\begin{aligned} 2 - 4 &= -2 \\ 0 - 2 &= -2 \\ -2 - 0 &= -2 \end{aligned}$$
The difference is the same every time, so the sequence is arithmetic with $d = -2$.
(A) is the first term. (B) is the right size with the wrong sign, from $4 - 2$. (D) is the change over two steps, $0 - 4$. QCAA did not publish how students answered in 2020.
Which option shows three consecutive terms for an arithmetic sequence?
Check the differences in each option.
$$\begin{aligned} &\text{(A)} \\ &{-7.0} - (-3.5) = -3.5 \\ &{-10.5} - (-7.0) = -3.5 \end{aligned}$$
Option (A) has a constant difference of $-3.5$, so it is arithmetic.
(B) multiplies by $-2$ each time, which is geometric. (C) has differences of $-10.5$ and then $+17.5$, so it is neither. (D) doubles each time, which is also geometric. Doubling feels like steady growth, and 12.6% chose (D). 76.5% of students chose (A).
The number of customers for a new gym’s first four days of operation is shown.
| Day | 1 | 2 | 3 | 4 |
| Customers | 6 | 18 | 30 | 42 |
a) Provide a justified decision for whether the number of customers forms an arithmetic sequence or a geometric sequence. [2 marks]
b) Assuming the sequence continues, display the number of customers for the gym’s first six days of operation in a scatterplot on the grid provided. [2 marks]
a) Find the differences between consecutive days:
$$42 - 30 = 30 - 18 = 18 - 6 = 12$$
The customer numbers form an arithmetic sequence with a common difference of 12.
1 mark · correctly identifies type of sequence
1 mark · correctly justifies type of sequence by identifying common difference of 12
The guide accepts “AP”, or circling ‘arithmetic’ in the stem, for the first mark. For the second it also accepts the $n$th term rule giving $d = 12$. The name and the evidence earn a mark each, so write both. A ratio check ($18 \div 6 = 3$, $30 \div 18 = 1.67$, not constant) rules out geometric, but the common difference is what earns the mark.
b) Continue the sequence: day 5 is $42 + 12 = 54$ and day 6 is $54 + 12 = 66$.
1 mark · correctly plots the number of customers for the first four days
1 mark · plots number of customers for 5th day and 6th day (follow-through allowed)
The 2025 report described this question among the ones students did well on.
Part 3
A recursive rule tells you how to get the next term from the one before. For an arithmetic sequence it is
$$\begin{aligned} t_{n+1} &= t_n + d \\ t_1 &= \text{the starting value} \end{aligned}$$
Read it as “next term equals this term plus $d$”. $t_{n+1} = t_n + 8$ with $t_1 = 25$ means start at 25 and add 8 each time: 25, 33, 41, 49, … The formula book does not print this form, so learn to read it on sight.
The recursive rule and the $n$th term rule describe the same sequence. $t_{n+1} = t_n + d$ with first term $t_1$ is $t_n = t_1 + (n-1)d$. Use the recursive rule to list a few terms, and the $n$th term rule to jump to a far-off one.
The syllabus asks you to show the terms in a table and on a graph. Three conventions decide the graph questions:
When a question wants a total, list the terms in a table and add them. 2024 Question 16b printed the table for you, and one of its two marks was for filling it in.
A swimmer has a weekly training routine to improve their fitness as modelled by the recursive function $T_{n+1} = T_n + 2$, where $T_n$ is the number of laps they swim in week $n$ and $T_1 = 4$. Which graph best represents the swimmer’s routine?
In each graph the points appear in week order, week 1 first, then the dashed line joins them.
Generate the terms. $T_1 = 4$, and each week adds 2 laps:
| Week $n$ | 1 | 2 | 3 | 4 | 5 | 6 |
| Laps $T_n$ | 4 | 6 | 8 | 10 | 12 | 14 |
The week number is the term number, so it goes on the horizontal axis, and the first point is (1, 4). Only (D) has both.
(A) has the right points with the axes swapped, and 16.0% chose it. (B) starts at week 0, treating 4 as $T_0$, and 13.9% chose it. (C) makes both slips. 63.6% of students chose (D).
The number of seats in each row of a theatre forms the terms of the arithmetic sequence $t_{n+1} = t_n + 8$, where $t_1 = 25$.
a) How many seats are in the second row of the theatre? [1 mark]
b) Complete the table and then calculate the total number of seats in the first four rows of the theatre. [2 marks]
| Row | 1 | 2 | 3 | 4 |
| Number of seats |
a) The rule says add 8 to get the next row:
$$t_2 = t_1 + 8 = 25 + 8 = 33$$
The second row of the theatre has 33 seats.
1 mark · correctly determines the number of seats in the second row
b) Keep adding 8:
$$\begin{aligned} t_3 &= t_2 + 8 = 41 \\ t_4 &= t_3 + 8 = 49 \end{aligned}$$
| Row | 1 | 2 | 3 | 4 |
| Number of seats | 25 | 33 | 41 | 49 |
1 mark · correctly completes the table to display the first four terms
$$25 + 33 + 41 + 49 = 148$$
There are 148 seats in the first four rows.
1 mark · calculates total number of seats in first four rows (follow-through allowed)
Row 1 is $t_1 = 25$. Starting the table at 33 treats 25 as a row 0 and gives a total of 180.
Try it yourself
Set the first term and the common difference. The graph, the table and both rules update together. Then choose how many terms to add, the way 2024 Paper 1 Question 16 asks for the seats in the first four rows.
Part 4
The $n$th term rule jumps straight to any term without listing the ones before it:
$$t_n = t_1 + (n-1)d$$
The $(n-1)$ counts gaps. To get from the 1st term to the 6th you take 5 steps of $d$, not 6. The same count works between any two terms: from $t_1$ to $t_3$ is $2d$, and from $t_4$ to $t_9$ is $5d$.
Marks are awarded for setting up before any arithmetic happens. In 2021 Paper 1 Question 20, two of the four marks are for identifying the model and its parameters. In 2020 Paper 2 Question 1, one mark is for defining the variables. The 2022, 2024 and 2025 reports all asked for the same habit. Every time:
QCAA’s guides say the parameter mark “may be implied by subsequent working”, but the model mark needs the model named. A bare $45\,100 - 2700 \times 9$ leaves the marker to infer both. Write them down and remove the doubt.
Determine the 6th term of the arithmetic sequence that begins 3, 9, …
$t_1 = 3$, $d = 9 - 3 = 6$ and $n = 6$.
$$\begin{aligned} t_6 &= 3 + (6-1) \times 6 \\ &= 33 \end{aligned}$$
(D) 729 treats the sequence as geometric, $3 \times 3^5$, and 15.1% chose it. The stem says arithmetic, so the multiplying is wrong even though 9 is $3 \times 3$. (A) 21 is $t_4$, two terms short. (C) 45 is $3 + 7 \times 6$, two terms too far. 75.4% of students chose (B).
Write the later term with the rule, substitute what you know, and solve for $d$. The number of gaps between the two terms is what multiplies $d$.
Exhibition organisers believe that the number of attendees increases each day as an arithmetic sequence. The organisers know that 353 people attended the first day and 439 people attended the third day.
a) Determine the common difference. [2 marks]
b) Use the result from 18a) to predict the number of people who will attend the sixth day. [2 marks]
a) Arithmetic sequence with $t_1 = 353$ and $t_3 = 439$. From the 1st day to the 3rd is two gaps:
$$\begin{aligned} t_3 &= t_1 + 2d \\ 439 &= 353 + 2d \\ 86 &= 2d \\ d &= 43 \end{aligned}$$
1 mark · correctly provides mathematical reasoning to support the answer
1 mark · correctly determines the common difference
The guide also accepts 86 where $n$ is defined as every 2 days. Dividing by 3, because there are three days, gives the wrong $d$.
b) $t_1 = 353$, $d = 43$, $n = 6$:
$$\begin{aligned} t_6 &= t_1 + 5d \\ &= 353 + 5 \times 43 \\ &= 568 \end{aligned}$$
They would expect 568 people to attend the sixth day.
1 mark · substitutes into an appropriate rule
1 mark · determines value
“Use the result from 18a)” signals follow-through: the guide allows FT marks in b), so a wrong $d$ substituted correctly still scores both marks. Recursion is accepted too: 353, 396, 439, 482, 525, 568. The 2020 report chose a sample that shows “translating the sixth day to meaning substitute $n = 6$ into the rule”.
The number of songs on a person’s playlist, n, in each week since joining a music streaming service, t, forms an arithmetic sequence, as shown by the graph.
Use the arithmetic sequence to predict the number of songs on this person’s playlist 25 weeks after joining the streaming service.
Read the first two points off the graph. The first term is the week 1 value:
$$t_1 = 87$$
1 mark · correctly determines $t_1$
$$\begin{aligned} d &= t_2 - t_1 \\ &= 209 - 87 \\ &= 122 \end{aligned}$$
1 mark · correctly determines $d$
Every gap on the graph is 122 ($331 - 209$, $453 - 331$, $575 - 453$), which confirms it.
$$\begin{aligned} t_n &= t_1 + (n-1)d \\ t_n &= 87 + 122(n-1) \end{aligned}$$
1 mark · uses an arithmetic sequence
$$\begin{aligned} t_{25} &= 87 + 122 \times 24 \\ &= 3015 \end{aligned}$$
The playlist is predicted to hold 3015 songs 25 weeks after joining.
1 mark · predicts number of songs at 25 weeks
Both $t_1$ and $d$ “may be implied by subsequent working”, and the last two marks allow follow-through. The guide also accepts a different $t_1$ “based on alternative definition of $n$”. If you count from week 0, the line passes through $87 - 122 = -35$ there, and $-35 + 122 \times 25$ gives the same 3015.
Notice the letters. The stem calls the number of songs n and the weeks t, the reverse of the formula book. QCAA’s own sample uses $t_n$ with $n$ as the week. Either works if you say what each letter means.
Try it yourself
Give two known terms and their positions. The number of steps between them divides the change, which gives $d$. Then predict any term. The two buttons load 2020 Paper 1 Question 18 and 2022 Paper 1 Question 18.
This $d$ is not a whole number. In a real count, check the two values you read.
Part 5
Most marks lost in this topic are lost by one term. The fix is one question: what real moment is $t_1$?
The check: substitute $n = 1$ into your rule and ask whether the result is true at the moment the sequence begins.
The counting rule: if the first term belongs to year $Y$, then year $Y + k$ is term $n = k + 1$. Count gaps, not terms. From 2012 to 2021 is 9 gaps, so $n = 10$.
Every arithmetic sequence is also a linear function $y = mx + c$, with $m = d$. The difference is where the counting starts. In $y = mx + c$, $x$ is the number of steps taken, starting from 0, and $c$ is the value before any step. In $t_n = t_1 + (n-1)d$, $n$ starts at 1. So $x = n - 1$.
The tractor is $n = 10$ as a sequence and $x = 9$ as a linear function, and both give $20 800. QCAA prints both methods in its marking guides with identical marks, so use whichever you count more reliably.
| Context | Start and step | Watch for |
|---|---|---|
| Simple interest investment | Start: the principal. Step: the interest earned each period. | Is $t_1$ the principal, or the balance after one period? Say which. |
| Taxi fare | Start: the flag fall. Step: the charge per kilometre. | A 1 km fare already includes one charge, so the 1 km fare is flag fall plus one rate. |
| Straight-line depreciation | Start: the purchase price. Step: minus the depreciation per year. | $d$ is negative. Count the years as gaps. |
| Draining or filling | Start: the starting volume. Step: litres per minute, negative if draining. | When $t_1$ is the starting volume, the time elapsed is $n - 1$. |
The syllabus names the first three contexts. In each one, say in words what $n$ counts.
A staircase is to be extended by installing n additional stairs. Each stair is 25 cm high. If the existing staircase reaches 1.2 m off the ground, which rule models the total height the stairs will reach in centimetres?
The answer is wanted in centimetres, so convert first: 1.2 m = 120 cm.
Each stair adds 25 cm, so $d = 25$.
$t_1$ is the height once the first additional stair is in place: $120 + 25 = 145$ cm.
$$t_n = 145 + (n-1) \times 25$$
Check $n = 1$: the rule gives 145 cm, one new stair on top of 120 cm, which is true. It also equals $120 + 25n$ for every $n$.
| Option | The mistake it rewards |
|---|---|
| (B) | Units never converted, and $t_1$ set at the existing height before any new stair is added. Two errors at once, and it drew 46.6% of the state. |
| (D) | Converted, then swapped the roles of $t_1$ and $d$, treating the one-off existing height as the repeating step. 31.0% chose it. |
| (A) | Roles swapped, and 1.45 is the height in metres after the first stair, left unconverted. 3.8% chose it. |
| (C) | Correct. Converts, then places $t_1$ after the first new stair. |
Only 17.5% chose (C), the lowest result of any multiple choice question QCAA published from 2021 to 2025. Two checks catch it. Units: if the stem mixes units, convert to the unit the answer asks for before modelling. Roles: ask which quantity happens once (it belongs inside $t_1$) and which repeats every term (that is $d$).
A farmer bought a tractor for $45 100 at the start of 2012. It depreciates by $2700 each year.
Identify and use a mathematical model to determine the value of the tractor at the start of 2021.
Option 1, arithmetic sequence. The start of 2012 is term 1, so the start of 2021 is term 10.
$$\begin{aligned} t_1 &= 45\,100 \\ d &= -2700 \\ n &= 10 \\ t_{10} &= ? \end{aligned}$$
1 mark · correctly identifies the model
1 mark · correctly identifies the parameters $t_1$, $d$ and $n$
$$\begin{aligned} t_n &= t_1 + (n-1)d \\ t_{10} &= 45\,100 - 2700(10-1) \\ &= 20\,800 \end{aligned}$$
1 mark · substitutes values into appropriate model
The tractor will be worth $20 800.
1 mark · determines value of tractor, including units
Option 2, linear function, earns the same four marks: $c = 45\,100$, $m = -2700$, $x = 9$, so $y = -2700 \times 9 + 45\,100 = 20\,800$. The guide also accepts modelling by recursion, nine subtractions of 2700.
The parameter mark may be implied by later working, and substitution and the final value allow follow-through. The last mark says “including units”, so the dollar sign matters. Using $n = 9$ in the sequence rule gives $23 500, and taking off ten years of depreciation gives $18 100. Both are off by one year.
Part 6
All 42 Paper 2 questions from 2020 to 2025 are single prompts with no parts. About three in four Paper 1 short responses do have parts (46 of 61). In Paper 2 you decide which steps to run and in what order, so practise sequence questions as one prompt.
When you know the value and want the term number, substitute $t_n$ and solve for $n$:
$$\begin{aligned} t_n &= t_1 + (n-1)d \\ n - 1 &= \frac{t_n - t_1}{d} \end{aligned}$$
Then decide what $n$ or $n - 1$ means in the context. If $t_1$ is the value at time zero, $n - 1$ is the time elapsed.
A water tank contains 12 500 L of water. The tap is accidently left on and the tank loses 135 L per minute. The tap is turned off when the tank has 5000 L of water left.
Use a mathematical model to determine how long the tap was left on to the nearest minute.
Option 1, arithmetic sequence. Define the variables: $n$ is the number of minutes starting at 1, and $t_n$ is the amount of water in the tank.
1 mark · correctly defines the variables
$$\begin{aligned} t_1 &= 12\,500 \\ d &= -135 \\ t_n &= 5000 \\ n &= ? \end{aligned}$$
1 mark · correctly identifies the parameters $t_1$, $d$ and $t_n$
$$\begin{aligned} t_n &= t_1 + (n-1)d \\ 5000 &= 12\,500 - 135(n-1) \\ 7500 &= 135(n-1) \end{aligned}$$
1 mark · substitutes values into appropriate model
$$\begin{aligned} n - 1 &= 55.5556 \\ n &= 56.5556 \end{aligned}$$
1 mark · determines $n$ value
$t_1$ is the full tank, at time zero, so the time the tap was on is $n - 1 = 55.56$ minutes. The tap was left on for about 56 minutes.
1 mark · states a reasonable answer rounded to the nearest minute
Option 2 is the linear function $y = -135x + 12\,500$, with $x$ the time the tap has been on. Setting $y = 5000$ gives $x = 55.5556$ directly, about 56 minutes, for the same five marks. The guide lists recursion as a third option, and it allows follow-through for substitution, $n$ and the final answer.
A quick check: after 55 minutes the tank holds 5075 L and after 56 minutes 4940 L, so 5000 L falls between them, nearer 56.
When two quantities each follow an arithmetic rule, they are equal where the rules give the same value. Set the rules equal and solve, or list both in a table until the values match. QCAA’s marking guide and the 2025 report accept either.
Fiore (F) and Goran (G) each hike for n hours along a track that joins a car park and a lookout. They both start their hike at 10 am, but from opposite ends of the track.
When Fiore and Goran meet on the track, they both take a 30-minute break. They then hike together to the car park, arriving at 4 pm.
Determine their hiking speed (km/h) from when Fiore and Goran finish their break to when they arrive at the car park.
Step 1 · Decide what $n = 0$ means. Here $n$ counts hours elapsed, so $n = 0$ is the start at 10 am, not $n = 1$. Check the models there:
$$\begin{aligned} F_0 &= 2 \times 0 = 0 \\ G_0 &= 12 - 3(0 - 1) = 15 \end{aligned}$$
Fiore starts at the car park and Goran starts 15 km away, so the track is 15 km long. QCAA’s sample response opens with the same substitution.
Step 2 · Find when and where they meet. Tabulate both models:
| $n$ | 0 | 1 | 2 | 3 |
| $F_n$ | 0 | 2 | 4 | 6 |
| $G_n$ | 15 | 12 | 9 | 6 |
Or solve them simultaneously:
$$\begin{aligned} 2n &= 12 - 3(n-1) \\ 2n &= 15 - 3n \\ 5n &= 15 \\ n &= 3 \end{aligned}$$
1 mark · correctly uses an appropriate method for determining when and where Fiore and Goran meet on the track
They meet 3 hours after the start of the hike, 6 km from the car park ($F_3 = G_3 = 6$).
1 mark · determines time since start of hike when Fiore and Goran meet on the track
1 mark · determines distance from car park where Fiore and Goran meet on the track
Step 3 · Account for the break. They meet at $10\text{ am} + 3\text{ h} = 1\text{ pm}$ and finish the break at 1:30 pm.
1 mark · determines time of day when Fiore and Goran finish their break
Step 4 · Speed. From 1:30 pm to 4 pm is 2.5 hours, and they still have the 6 km back to the car park.
$$s = \frac{d}{t} = \frac{6\text{ km}}{2.5\text{ h}} = 2.4\text{ km/h}$$
Their hiking speed from finishing the break to arriving at the car park is 2.4 km/h.
1 mark · determines hiking speed
1 mark · shows logical organisation
The guide defines logical organisation as “clear flow of the solution using appropriate mathematical terminology, symbols, conventions and representations”. Labelled steps, stated rules and a closing sentence with units earn it. The 2025 report printed two student methods, one solving simultaneously and one finding equal terms from each model.
Four more Paper 2 questions use an arithmetic sequence as one stage. Each is worked in full in the guide for its main stage.
| Question | The arithmetic stage | Worked in |
|---|---|---|
| 2021 Q4 (6 marks) | A linear model that grows by $126 a year, compared with a compound model read from a graph. | Geometric sequences |
| 2022 Q6 (7 marks) | The $y$-intercepts of a pattern of lines go up by 1.5 each time from 1.2, so the fifth is $1.2 + 4 \times 1.5 = 7.2$. The slopes form a geometric sequence. | Geometric sequences |
| 2023 Q5 (7 marks) | Road lengths from $t_n = t_1 + 2(n-1)$ with $t_1 = 1$, so $t_2 = 3$, $t_3 = 5$, $2t_1 = 2$, $2t_2 = 6$ and $t_1 + t_2 = 4$. One mark is for “correctly applies arithmetic sequence to determine all road lengths”, then a Hamiltonian cycle and the time. | Graphs and networks |
| 2024 Q6 (6 marks) | Daily meals $m_n = m_1 + 3(n-1)$ with $m_1 = 60$, so $m_5 = 72$ in 2025. Accommodation is geometric, and the 7-day total of about $1734 is tested against an estimate of $1500 to $2000. | Geometric sequences |
Part 7
The marking guides give one mark per dot point, and in this topic the dot points are mostly the setup: the model, the variables, the parameters and the substitution. Here is what they reward, in their words.
| What earns the mark | Where the guides say so |
|---|---|
| Naming the model | “correctly identifies the model” (2021 P1 Q20). In 2025 P1 Q21 “correctly identifies type of sequence” is a mark on its own. |
| Defining the variables | “correctly defines the variables” (2020 P2 Q1). The 2022, 2024 and 2025 reports all asked for it. |
| The parameters | “correctly identifies the parameters $t_1$, $d$ and $n$” (2021 P1 Q20), “correctly determines $t_1$” and “correctly determines $d$” (2022 P1 Q18). These may be implied by later working. |
| Substitution | “substitutes values into appropriate model” (2020 P2 Q1, 2021 P1 Q20), with follow-through from earlier errors. |
| The answer, in context | “determines value of tractor, including units” (2021). “states a reasonable answer rounded to the nearest minute” (2020 P2 Q1). |
| Logical organisation | One mark in 2025 P2 Q6, for a clear flow using appropriate terminology, symbols and conventions. |
Every multi-step arithmetic question in the guides allows follow-through. A wrong $d$ substituted correctly still earns the substitution and answer marks. So never leave part b) blank because part a) went wrong, and never rub out a wrong value you have used.
| Checkpoint | How to run it |
|---|---|
| Substitute $n = 1$ | Does your rule give the value that is true when the sequence starts? This is the check that settles the staircase question. |
| Sign of $d$ | Depreciation, draining and cooling give a negative $d$. A positive $d$ next to a falling quantity means something has gone wrong. |
| Unit sweep | Two units in the stem mean a conversion. Check which unit the answer should use. |
| Gaps, not terms | From $t_1$ to $t_3$ is $2d$. From 2012 to 2021 is 9 gaps, so $n = 10$. |
| Arithmetic or geometric? | Constant differences mean arithmetic, constant ratios mean geometric. 2025 Paper 1 Q21 gave one mark for naming the sequence and one for justifying it with the common difference of 12. Name it and show the evidence. |
| Answer the question asked | “How long was the tap left on” wants a time in minutes, not a term number. Re-read the last line of the stem before you write the answer. |
Game plan
Run these steps on any arithmetic sequence question, in either paper.
Practice
I wrote these four to the QCAA format and checked every answer in Python. The marking shown is my suggested marking, in the style of the QCAA guides. Question 4 has the shape of 2025 Paper 2 Question 6, so attempt it as one question with no parts and give yourself about 14 minutes, the Paper 2 rate of 90 minutes for 38 marks.
A rainwater tank holds 240 L and is drained at 8 L per minute. Which rule models the volume, in litres, remaining after n minutes of draining?
The stem defines $t_n$ as the volume after $n$ minutes, so the rule must give the volume after 1 minute at $n = 1$: $240 - 8 = 232$ L.
$$\begin{aligned} t_n &= 232 - (n-1) \times 8 \\ &= 240 - 8n \end{aligned}$$
(C) gives 240 L at $n = 1$, but after 1 minute the tank holds 232 L. Starting from $t_1 = 240$ is fine when $n$ is just a term number, as in 2020 Paper 2 Question 1, but this stem fixes $n$ as minutes of draining. (A) also adds instead of subtracting. (D) swaps the roles of $t_1$ and $d$.
A courier charges a flat pick-up fee plus a fixed amount per kilometre. A 4 km delivery costs $23.00 and a 9 km delivery costs $35.50.
a) Determine the charge per kilometre. [2 marks]
b) Use the result from a) to determine the cost of a 15 km delivery. [2 marks]
a) Arithmetic sequence, with $t_n$ the cost of an $n$ km delivery: $t_4 = 23.00$ and $t_9 = 35.50$. From 4 km to 9 km is five gaps:
$$\begin{aligned} t_9 &= t_4 + 5d \\ 35.50 &= 23.00 + 5d \\ 5d &= 12.50 \\ d &= 2.50 \end{aligned}$$
1 mark · provides mathematical reasoning, using five gaps between the two deliveries
1 mark · determines the charge of $2.50 per kilometre
b) From 4 km to 15 km is eleven gaps:
$$\begin{aligned} t_{15} &= t_4 + 11d \\ &= 23.00 + 11 \times 2.50 \\ &= 50.50 \end{aligned}$$
1 mark · substitutes into an appropriate rule (follow-through allowed)
1 mark · determines the cost of $50.50
Another route: the flat fee is $23.00 - 4 \times 2.50 = 13.00$, and $13.00 + 15 \times 2.50 = 50.50$ gives the same $50.50. Counting four gaps instead of five gives $3.125 per km, which is the slip this question is built to catch.
The number of tiles in each layer of a display forms the arithmetic sequence $t_{n+1} = t_n + 6$, where $t_1 = 14$.
a) How many tiles are in the fourth layer? [1 mark]
b) Complete the table and calculate the total number of tiles in the first five layers. [2 marks]
| Layer | 1 | 2 | 3 | 4 | 5 |
| Number of tiles |
a) $t_1 = 14$ and $d = 6$:
$$t_4 = 14 + (4-1) \times 6 = 32$$
There are 32 tiles in the fourth layer.
1 mark · correctly determines the number of tiles in the fourth layer
| Layer | 1 | 2 | 3 | 4 | 5 |
| Number of tiles | 14 | 20 | 26 | 32 | 38 |
1 mark · correctly completes the table to display the first five terms
$$\begin{aligned} &14 + 20 + 26 + 32 + 38 \\ &= 130 \end{aligned}$$
There are 130 tiles in the first five layers.
1 mark · calculates the total number of tiles (follow-through allowed)
There is no sum formula in this course, so the table is the intended method, exactly as in 2024 Paper 1 Question 16.
A council is installing street lights along a new 1.8 km road that runs from a roundabout to a bridge. Two crews work at the same time, each installing one light every 12 minutes, both starting at 7:00 am.
Both crews stop once their next light would be positioned past a light the other crew has already installed.
Ciara claims the crews will have finished installing all the lights before 11:00 am. Evaluate the reasonableness of Ciara’s claim.
Nothing in the stem tells you to compare $A_n$ with $B_n$. That comparison is the step to find. The crews install their $n$th lights at the same time, so compare the two sequences light by light.
$$\begin{aligned} 45n &= 1800 - 60n \\ 105n &= 1800 \\ n &\approx 17.14 \end{aligned}$$
1 mark · compares the two sequences to locate where the crews meet
So the crews pass each other between the 17th and 18th lights. Check the terms around it:
| $n$ | $A_n$ (m) | $B_n$ (m) | Gap (m) |
|---|---|---|---|
| 16 | 720 | 840 | 120 |
| 17 | 765 | 780 | 15 |
| 18 | 810 | 720 | A would pass B |
An 18th light from Crew A would sit at 810 m, past Crew B’s light at 780 m, and Crew B’s 18th at 720 m would be past Crew A’s at 765 m. Both crews stop after 17 lights each.
1 mark · determines that each crew installs 17 lights
$$\begin{aligned} 17 \times 12 &= 204\text{ minutes} \\ &= 3\text{ h } 24\text{ min} \end{aligned}$$
1 mark · determines the time taken
Starting at 7:00 am, they finish at 10:24 am.
1 mark · determines the finishing time
10:24 am is before 11:00 am, with 36 minutes to spare, so Ciara’s claim is reasonable.
1 mark · compares the finishing time with 11:00 am and states a judgement on Ciara’s claim in context
1 mark · shows logical organisation
If you take the first light as installed at 7:00 am rather than 7:12 am, they finish at 10:12 am. The verdict is the same. A correct time with no comparison against 11:00 am and no judgement leaves the verdict mark unscored.
Sources: QCAA General Mathematics external assessment question and response books, multiple choice books and marking guides, 2020 to 2025. General Mathematics 2025 v1.3 general senior syllabus, Unit 3 Topic 4, and the 2019 v1.2 syllabus for comparison. QCAA General Mathematics subject reports, 2020 to 2025. General Mathematics formula book, and the formula sheet used for the 2020 to 2024 papers. Every number on this page was recomputed in Python.
Past questions are from the 2020–2025 QCAA General Mathematics external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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