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

Practise 36 exam-style questions on nth term of linear and quadratic sequences for Edexcel GCSE Maths (Non-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.

36 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.
5, 9, 13, 17, 21

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

Model answer

Common difference \(= 4\), so the \(n\)th term contains \(4n\).
\(4n\): 4, 8, 12, 16, 20
Each term is 1 more than \(4n\).
\(n\)th term \(= 4n + 1\)

Explanation

The terms go up by 4, so compare the sequence with the 4 times table, \(4n\). Every term is 1 more than the matching multiple of 4, giving \(4n + 1\). Check with \(n = 5\): \(4 \times 5 + 1 = 21\).

Mark scheme

1 mark
M1 for \(4n + k\) where \(k\) is any number (including \(k = 0\))
or a common difference of 4 linked to \(4n\); correct answer with no working scores full marks
1 mark
A1 for \(4n + 1\)
oe; \(n + 4\) scores 0

Question 2

2 marks

Here are the first five terms of a quadratic sequence.
3, 12, 27, 48, 75

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

Model answer

First differences: 9, 15, 21, 27
Second difference: 6, so the \(n\)th term contains \(3n^2\).
\(3n^2\): 3, 12, 27, 48, 75, which are exactly the terms.
\(n\)th term \(= 3n^2\)

Explanation

Half of the second difference of 6 gives the coefficient 3 of \(n^2\). Subtracting \(3n^2\) leaves 0 every time, so there is no linear part and no constant. You might also notice the terms are 3 times 1, 4, 9, 16, 25.

Mark scheme

1 mark
M1 for a second difference of 6, or for \(3n^2 + bn + c\) with \(b\) and \(c\) any numbers
or for noticing that the terms are 3 times the square numbers; correct answer with no working scores full marks
1 mark
A1 for \(3n^2\)
oe

Question 3

3 marks

Here are the first five terms of a sequence.
2, 8, 18, 32, 50

Ali says, "150 is a term in this sequence."

Is Ali correct?
You must show how you get your answer.

Model answer

First differences: 6, 10, 14, 18; second difference 4, so the \(n\)th term contains \(2n^2\).
\(2n^2\): 2, 8, 18, 32, 50, so the \(n\)th term is \(2n^2\).
\(2n^2 = 150\)
\(n^2 = 75\)
75 is not a square number (\(8^2 = 64\), \(9^2 = 81\)), so \(n\) is not a whole number.
Ali is not correct.

Explanation

The terms are twice the square numbers, so the \(n\)th term is \(2n^2\). A number is a term only if solving gives a positive whole number \(n\). Since 75 is not a perfect square, 150 is not a term: it falls between the 8th term, 128, and the 9th term, 162.

Mark scheme

1 mark
B1 for the \(n\)th term \(2n^2\)
oe
1 mark
M1 for \(2n^2 = 150\) leading to \(n^2 = 75\), or working out the 8th term (128) and the 9th term (162)
ft their quadratic \(n\)th term
1 mark
C1 for No, with a correct reason, e.g. 75 is not a square number so \(n\) is not a whole number, or 150 lies between the 8th and 9th terms

Question 4

4 marks

The \(n\)th term of a sequence is \(an^2 + bn\), where \(a\) and \(b\) are integers.
The 2nd term of the sequence is 2.
The 5th term of the sequence is 35.

Find the value of \(a\) and the value of \(b\).

Model answer

2nd term: \(4a + 2b = 2\), so \(2a + b = 1\)
5th term: \(25a + 5b = 35\), so \(5a + b = 7\)
Subtract: \(3a = 6\), so \(a = 2\)
\(2(2) + b = 1\), so \(b = -3\)
Check: \(2(25) - 3(5) = 50 - 15 = 35\)

Explanation

Substitute \(n = 2\) and \(n = 5\) into \(an^2 + bn\) to get two simultaneous equations. Dividing each equation by a common factor first makes the elimination easier. Check the answer with both given terms.

Mark scheme

1 mark
M1 for \(4a + 2b = 2\)
oe; correct answer with no working scores full marks
1 mark
M1 for \(25a + 5b = 35\)
oe
1 mark
M1 for a correct method to eliminate \(a\) or \(b\), e.g. \(b = 1 - 2a\) substituted, or matching coefficients
ft their equations
1 mark
A1 for \(a = 2\) and \(b = -3\)

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.