nth term of linear and quadratic sequences - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 30 exam-style questions on nth term of linear and quadratic sequences for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

30 questions
1–5 marks each
Instant marking
No sign-up needed

20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Here are the first five terms of an arithmetic sequence.

2.4, 3.1, 3.8, 4.5, 5.2

Find an expression, in terms of \(n\), for the \(n\)th term of the sequence.

Model answer

Common difference \(= 3.1 - 2.4 = 0.7\)
Term before the first term \(= 2.4 - 0.7 = 1.7\)
\(n\)th term \(= 0.7n + 1.7\)

Explanation

In an arithmetic sequence the \(n\)th term is (common difference)\( \times n\) + (the term before the first term). Check your expression: when \(n = 5\), \(0.7 \times 5 + 1.7 = 5.2\), which matches the 5th term.

Mark scheme

1 mark
M1 for \(0.7n + k\) where \(k\) is a number, or for \(0.7n\) seen
k may be 0; correct answer with no working scores full marks
1 mark
A1 for \(0.7n + 1.7\)
oe, e.g. \(1.7 + 0.7n\); exact values

Question 2

3 marks

A taxi company charges a fixed amount plus the same amount for each mile.

The cost of a 1-mile journey is £3.80
The cost of a 2-mile journey is £5.45
The cost of a 3-mile journey is £7.10
The cost of a 4-mile journey is £8.75

Find an expression, in terms of \(n\), for the cost in pounds of a journey of \(n\) miles, and use it to work out the cost of a 13-mile journey.

Model answer

Increase per mile \(= 5.45 - 3.80 = 1.65\)
Fixed charge \(= 3.80 - 1.65 = 2.15\)
Cost \(= 1.65n + 2.15\) pounds
13 miles: \(1.65 \times 13 + 2.15 = 21.45 + 2.15 = 23.60\)
The cost is £23.60

Explanation

The costs form an arithmetic sequence: they go up by £1.65 each mile, and the fixed charge is the cost of "0 miles", one step back from the first term. Substituting \(n = 13\) is quicker and more reliable than continuing the sequence term by term.

Mark scheme

1 mark
M1 for \(1.65n + k\), or for 1.65 identified as the cost per mile
correct answer with no working scores full marks
1 mark
A1 for \(1.65n + 2.15\)
oe; exact values
1 mark
B1 for £23.60
ft their linear expression; accept £23.6

Question 3

3 marks

Here are the first five terms of a quadratic sequence.

6, 15, 28, 45, 66

Find an expression for the \(n\)th term of the sequence and use it to show that no term of the sequence is a prime number.

Model answer

First differences: 9, 13, 17, 21
Second differences: 4, 4, 4, so the \(n^2\) term is \(2n^2\)
Terms \(-\, 2n^2\): 4, 7, 10, 13, 16, which is \(3n + 1\)
\(n\)th term \(= 2n^2 + 3n + 1 = (2n + 1)(n + 1)\)
For \(n \ge 1\), \(2n + 1 \ge 3\) and \(n + 1 \ge 2\), so every term is the product of two whole numbers that are both greater than 1.
So every term has a factor other than 1 and itself, and no term is prime.

Explanation

Find the \(n\)th term in the usual way, then factorise it. A prime number has exactly two factors, 1 and itself, so showing every term is a product of two whole numbers each bigger than 1 proves none is prime. Saying "the first five terms are not prime" is not enough: you must show it for every term.

Mark scheme

1 mark
M1 for \(2n^2\) seen, or second differences of 4 and a correct method to find the linear part
1 mark
A1 for \(2n^2 + 3n + 1\)
oe
1 mark
C1 for \((2n + 1)(n + 1)\) with a correct explanation, e.g. both factors are whole numbers greater than 1 (since \(n \ge 1\)), so every term has a factor other than 1 and itself
dependent on A1; answer only scores 0 for this mark

Question 4

5 marks

Here are the first five terms of a quadratic sequence.

2.5, 7, 13.5, 22, 32.5

Is 2214 a term of this sequence?
You must show how you get your answer.

Model answer

First differences: 4.5, 6.5, 8.5, 10.5
Second differences: 2, 2, 2, so the \(n^2\) term is \(n^2\)
Terms \(-\, n^2\): 1.5, 3, 4.5, 6, 7.5, which is \(1.5n\)
\(n\)th term \(= n^2 + 1.5n\)
\(n^2 + 1.5n - 2214 = 0\)
\(n = \dfrac{-1.5 + \sqrt{2.25 + 8856}}{2} = \dfrac{-1.5 + 94.118...}{2} = 46.30...\)
46th term \(= 2116 + 69 = 2185\); 47th term \(= 2209 + 70.5 = 2279.5\)
\(n\) is not a whole number, so 2214 is not a term of the sequence.

Explanation

First find the \(n\)th term from the differences. Then set it equal to 2214 and solve with the quadratic formula: 2214 is a term only if \(n\) is a positive whole number. Evaluating the 46th and 47th terms shows that 2214 falls between two terms.

Mark scheme

1 mark
M1 for second differences of 2 leading to \(n^2\)
1 mark
M1 for subtracting \(n^2\) to get 1.5, 3, 4.5, 6, 7.5 or for \(1.5n\) seen
ft their an²
1 mark
A1 for \(n^2 + 1.5n\)
oe
1 mark
M1 for \(n^2 + 1.5n = 2214\) and a correct method to solve, e.g. \(n = \dfrac{-1.5 + \sqrt{1.5^2 + 4 \times 2214}}{2}\), or for working out the 46th and 47th terms
ft their quadratic nth term
1 mark
C1 for No, with \(n = 46.3...\) (not a whole number) or with 46th term = 2185 and 47th term = 2279.5
dependent on all previous marks

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

AI Tutor Hints first, answers after you try

Hi! Stuck? Ask me for a hint and I’ll help you work it out. Once you’ve checked your answer I can explain exactly where the marks are.