What is A24: Special sequences?

Square and cube numbers have forms n^2 and n^3, while triangular numbers have form n(n+1)/2. An arithmetic progression has a constant first difference and a quadratic sequence has constant second differences.

How many marks are A24 questions worth?

In verified past-paper appearances, A24 questions on Edexcel GCSE Maths 1MA1 have been worth 2–7 marks.

Which past exam papers has A24 appeared in?

A24 has appeared in 10 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A24

Recognise and use triangular, square and cube numbers, arithmetic progressions, Fibonacci type sequences, quadratic sequences, simple geometric progressions (r^n, r rational > 0 or a surd) and others

Special sequences

Worked answers, methods and verified real exam appearances for A24 on Edexcel GCSE Maths 1MA1.

Explanation

Square and cube numbers have forms $n^2$ and $n^3$ , while triangular numbers have form $\dfrac{n(n+1)}{2}$ . An arithmetic progression has a constant first difference and a quadratic sequence has constant second differences.
Fibonacci-type sequences form later terms from preceding terms.
At Higher tier, a geometric progression has a constant positive rational or surd multiplier and can be written using powers such as $r^n$ .
Check several consecutive steps before identifying a sequence, then use the defining rule consistently.
The examiner expects the named structure to be supported by differences, ratios or a valid term relationship rather than by visual resemblance alone.

Worked example

Higher tier: The sequence $2,2\sqrt3,6,6\sqrt3,\ldots$ is geometric. Find its common ratio and eighth term.

1.Divide consecutive terms: $2\sqrt3\div2=\sqrt3$ , so $r=\sqrt3$ .
2.Use $u_n=2(\sqrt3)^{n-1}$ .
3. $u_8=2(\sqrt3)^7=2(27\sqrt3)=54\sqrt3$ .

Answer: Common ratio $\sqrt3$ ; eighth term $54\sqrt3$ .

Common mistakes

Don't call a sequence arithmetic after checking only one pair of terms.
Don't use first differences to identify a quadratic sequence instead of checking that second differences are constant.
Don't replace the exact surd ratio by a decimal and lose the exact form.

Exam tip

Show a short difference table or two equal consecutive ratios to justify the sequence type.

Worked practice

Tier 1 · Easy

The sequence

1,4,9,16,\ldots

is made from a named type of number. Name the type and write down the next two terms.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Square numbers $25,36$	2	The terms are $1^2,2^2,3^2,4^2$ . Therefore they are square numbers and the next terms are $5^2=25$ and $6^2=36$ .

Tier 2 · Standard

A Fibonacci-type sequence begins

3,7,10,17,27,\ldots

, with each term after the second equal to the sum of the previous two. Find the eighth term.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$115$	3	Continue the rule: the sixth term is $17+27=44$ , the seventh is $27+44=71$ , and the eighth is $44+71=115$ .

Tier 3 · Hard

Higher only: A geometric progression begins

2,2\sqrt3,6,6\sqrt3,\ldots

. State the common ratio and find the eighth term in exact form.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Common ratio $=\sqrt3$ Eighth term $=54\sqrt3$	4	Each term is multiplied by $\sqrt3$ . The $n$ th term is $2(\sqrt3)^{n-1}$ . For $n=8$ , this is $2(\sqrt3)^7=2(27\sqrt3)=54\sqrt3$ .

Tier 1 · Easy

Name the special numbers shown by

1,3,6,10,15,\ldots

and give the following term.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	Triangular numbers $21$	2	These are triangular numbers. Their differences are $2,3,4,5$ , so the next difference is $6$ and the next term is $15+6=21$ .

Tier 2 · Standard

The fifth term of an arithmetic sequence is

18

. The difference between consecutive terms is

4

. Work out the first term and the ninth term.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	First term $=2$ Ninth term $=34$	3	The fifth term is four common differences after the first, so the first term is $18-4(4)=2$ . The ninth term is eight common differences after the first, giving $2+8(4)=34$ .

Tier 3 · Hard

In a Fibonacci-type sequence, each term after the second is the sum of the previous two. The fourth term is

26

and the fifth term is

43

. Work backwards to find the first three terms.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$8,9,17$	3	The third term is $43-26=17$ . The second term is $26-17=9$ . The first term is $17-9=8$ , so the first three terms are $8,9,17$ .

Tier 2 · Standard

The sequence begins

3,8,15,24,35

. Explain why this is a quadratic sequence. Work out the next two terms.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	The second differences are constant and equal to $2$ $48,63$	3	The first differences are $5,7,9,11$ , so the second differences are all $2$ and the sequence is quadratic. Continue the odd-number first differences with $13$ and $15$ : $35+13=48$ and $48+15=63$ .

Tier 3 · Hard

Find the smallest integer greater than

100

that is both a square number and a cube number. Show enough working to justify that it is the smallest.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$729$	4	The cube numbers greater than $100$ begin $5^3=125$ , $6^3=216$ , $7^3=343$ , $8^3=512$ and $9^3=729$ . The first four are not square numbers, while $729=27^2$ as well as $9^3$ . Therefore $729$ is the smallest integer greater than $100$ that is both types.

Tier 3 · Hard

The sequence

1,1,4,8,9,27,16,64,\ldots

is formed by interleaving two named sequences. Identify the two sequences and work out the next four terms.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	The odd-position terms are square numbers and the even-position terms are cube numbers $25,125,36,216$	4	The odd-position terms are $1^2,2^2,3^2,4^2,\ldots$ and the even-position terms are $1^3,2^3,3^3,4^3,\ldots$ . The fifth square and fifth cube are $25$ and $125$ , followed by the sixth square and sixth cube, $36$ and $216$ .

Q10

Tier 3 · Hard

A geometric progression has positive common ratio. Its second term is

12

and its fifth term is

81/2

. Work out the common ratio, the first term and the seventh term. Give exact answers.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Common ratio $=\dfrac32$ First term $=8$ Seventh term $=\dfrac{729}{8}$	4	From the second to the fifth term there are three multiplications by the ratio, so $r^3=(81/2)/12=27/8$ . The positive ratio is $r=3/2$ . The first term is $12\div(3/2)=8$ . The seventh term is $(81/2)(3/2)^2=729/8$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2021-11	1F	Q13	3	Non-calculator	Foundation	QP MS
2023-06	3H	Q13	3	Allowed	Higher	QP MS
2019-06	2F	Q28	3	Allowed	Foundation	QP MS
2022-11	1H	Q23	5	Non-calculator	Higher	QP MS
2023-11	3F	Q8	4	Allowed	Foundation	QP MS
2021-11	3F	Q25	3	Allowed	Foundation	QP MS
2021-11	3H	Q5	3	Allowed	Higher	QP MS
2019-11	3F	Q8	2	Allowed	Foundation	QP MS
2023-11	1H	Q22	4	Non-calculator	Higher	QP MS
2024-06	3H	Q20	7	Allowed	Higher	QP MS

Other points in A Algebra · sequences and functions

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