What is A25: nth term?

For a linear sequence with common difference d, start with dn and adjust the constant so the expression gives the first term when n=1.

How many marks are A25 questions worth?

In verified past-paper appearances, A25 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has A25 appeared in?

A25 has appeared in 11 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A25

Deduce expressions to calculate the nth term of linear and quadratic sequences

nth term

Worked answers, methods and verified real exam appearances for A25 on Edexcel GCSE Maths 1MA1.

Explanation

For a linear sequence with common difference $d$ , start with $dn$ and adjust the constant so the expression gives the first term when $n=1$ .
At Higher tier, a quadratic sequence $an^2+bn+c$ has constant second difference $2a$ .
Find $a$ , subtract the values of $an^2$ from the original terms, and identify the linear rule left behind to obtain $b$ and $c$ .
The examiner expects an expression in $n$ , not merely the next term.
Verify the rule against at least two supplied terms; this detects a shifted index or an incorrect constant before the final answer.

Worked example

Higher tier: Find the $n$ th term of $2,7,14,23,34,\ldots$ .

1.First differences are $5,7,9,11$ , so the second difference is $2$ and $a=1$ .
2.Subtract $n^2$ from the terms to get $1,3,5,7,9$ .
3.The remainder has rule $2n-1$ , so combine the parts.

Answer: $n^2+2n-1$ .

Common mistakes

Don't use the constant second difference as $a$ instead of recognising it is $2a$ .
Don't use the first term itself as the constant in a linear $n$ th-term rule.
Don't find a rule that matches one term and fail to verify it against later terms.

Exam tip

For a quadratic sequence, display the first and second differences because they secure the method for finding the $n^2$ coefficient.

Worked practice

Tier 1 · Easy

Find an expression for the

n

th term of

7,11,15,19,\ldots

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$4n+3$	2	The common difference is $4$ , so start with $4n$ . This gives $4$ when $n=1$ , which is $3$ below the first term $7$ . Therefore the rule is $4n+3$ .

Tier 2 · Standard

Higher only: Find an expression for the

n

th term of

2,7,14,23,34,\ldots

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$n^2+2n-1$	4	The first differences are $5,7,9,11$ , so the second difference is $2$ and the quadratic begins with $n^2$ . Subtracting $n^2$ gives $1,3,5,7,9$ , whose rule is $2n-1$ . Therefore the $n$ th term is $n^2+2n-1$ .

Tier 3 · Hard

Higher only: A quadratic sequence starts

4,15,32,55

. Deduce its

n

th term and determine the position of the term equal to

207

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$3n^2+2n-1$ $207$ is the eighth term	6	The first differences are $11,17,23$ , so the second difference is $6$ and $a=3$ . Subtracting $3n^2$ from the terms gives $1,3,5,7$ , whose rule is $2n-1$ . Hence the term is $3n^2+2n-1$ . Set this equal to $207$ : $3n^2+2n-208=0=(3n+26)(n-8)$ . The positive integer solution is $n=8$ .

Tier 1 · Easy

Find an expression for the

n

th term of

20,17,14,11,\ldots

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$23-3n$ (or $-3n+23$ )	2	The common difference is $-3$ , so start with $-3n$ . This gives $-3$ when $n=1$ , which is $23$ below the first term $20$ . Therefore the rule is $-3n+23$ , equivalently $23-3n$ .

Tier 2 · Standard

A linear sequence has sixth term

29

and fourteenth term

77

. Find an expression for its

n

th term.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$6n-7$	3	There are eight equal steps from the sixth to the fourteenth term, and the total increase is $77-29=48$ . The common difference is therefore $48/8=6$ . A rule $6n+c$ gives $29$ when $n=6$ , so $36+c=29$ and $c=-7$ . The $n$ th term is $6n-7$ .

Tier 3 · Hard

Find an expression for the

n

th term of

13,20,27,34,\ldots

. Work out the

30

th term. Is

200

a term of this sequence? Give a reason.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$7n+6$ The $30$ th term is $216$ $200$ is not a term of the sequence	4	The common difference is $7$ , so the $n$ th term is $7n+6$ . The $30$ th term is $7(30)+6=216$ . If $200$ were a term, then $7n+6=200$ , giving $n=194/7=27\dfrac57$ . Since a term position must be a positive integer, $200$ is not in the sequence.

Tier 2 · Standard

The

n

th term of a linear sequence is

7n+k

. The

12

th term is

79

. Work out the value of

k

and write down the first three terms of the sequence.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$k=-5$ $2,9,16$	3	Substitute $n=12$ : $7(12)+k=79$ , so $84+k=79$ and $k=-5$ . The rule is $7n-5$ , which gives $2,9,16$ for $n=1,2,3$ .

Tier 3 · Hard

Higher only: A quadratic sequence has

n

th term

n^2+bn+c

. Its second term is

9

and its sixth term is

57

. Work out

b

and

c

. Hence write down the

n

th term and work out the

10

th term.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$b=4$ , $c=-3$ $n^2+4n-3$ The $10$ th term is $137$	4	Using the second term gives $4+2b+c=9$ , so $2b+c=5$ . Using the sixth gives $36+6b+c=57$ , so $6b+c=21$ . Subtracting gives $4b=16$ , hence $b=4$ and then $c=-3$ . The rule is $n^2+4n-3$ , and its $10$ th term is $100+40-3=137$ .

Tier 3 · Hard

Higher only: The quadratic sequence begins

10,17,20,19,14,\ldots

. Find an expression for its

n

th term. Hence work out the greatest term in the sequence and its position.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$-2n^2+13n-1$ The greatest term is $20$ , at position $n=3$	5	The first differences are $7,3,-1,-5$ , so the second difference is $-4$ and the rule begins with $-2n^2$ . Adding $2n^2$ to the terms gives $12,25,38,51,64$ , whose rule is $13n-1$ . Hence $u_n=-2n^2+13n-1$ . Also $u_{n+1}-u_n=11-4n$ , which is positive up to $n=2$ and negative from $n=3$ , so the maximum is $u_3=20$ .

Q10

Tier 3 · Hard

Sequence A begins

5,12,19,26,\ldots

and sequence B begins

104,100,96,92,\ldots

. Find an expression for the

n

th term of each sequence. Hence find the first position at which the two sequences have the same term, and give the value of that term.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Sequence A: $7n-2$ Sequence B: $108-4n$ They are equal at the $10$ th term, which is $68$	5	Sequence A has common difference $7$ , so its rule is $7n-2$ . Sequence B has common difference $-4$ , so its rule is $108-4n$ . Equating them gives $7n-2=108-4n$ , hence $11n=110$ and $n=10$ . Substitution gives $7(10)-2=68$ , also $108-4(10)=68$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-06	1F	Q19	2	Non-calculator	Foundation	QP MS
2024-06	1F	Q20	2	Non-calculator	Foundation	QP MS
2024-06	1H	Q1	2	Non-calculator	Higher	QP MS
2023-06	2H	Q15	3	Allowed	Higher	QP MS
2024-11	2H	Q15	6	Allowed	Higher	QP MS
2019-11	3H	Q20	2	Allowed	Higher	QP MS
2022-11	2F	Q20	4	Allowed	Foundation	QP MS
2022-11	2H	Q3	4	Allowed	Higher	QP MS
2019-06	3H	Q16	3	Allowed	Higher	QP MS
2022-06	3F	Q8	2	Allowed	Foundation	QP MS
2023-11	3H	Q14	2	Allowed	Higher	QP MS

Other points in A Algebra · sequences and functions

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