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Quadratic sequences

Free Printable Quadratic sequences PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 8+)

Download free quadratic sequences worksheets (PDF), GCSE maths past paper questions and worked solutions for Quadratic sequences (8F). This is a Grade 8+ algebra topic from the GCSE questions by topic collection for Foundation and Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Corbett Maths, Maths Made Easy, Naiker Maths, 1st Class Maths, Metatutor, Dr Austin, GCSE Maths Rocket, Maths4EveryOne, IGCSE MT, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our quadratic sequences revision notes and worked examples with diagrams, try real GCSE past paper quadratic sequences questions with step-by-step solutions, or download a PDF of all quadratic sequences worksheet links.

Free Quadratic sequences PDF Worksheets & Answers

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GCSE Maths Grade 8+ Revision Notes & Method

Guide to Quadratic sequences: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

A quadratic sequence has nth term an2 + bn + c and a constant second difference. Work out the first differences and second differences between terms, then halve the second difference to find a (since 2a = 2nd difference).

Write out the values of an2 for n = 1, 2, 3, 4 and subtract them from the original sequence. The remaining numbers form a simple linear sequence bn + c; find its nth term and combine with an2.

Example 1: Find the nth term of 4, 11, 22, 37, 56, ...

Solution:

1st diffs: +7, +11, +15, +19 ⇒ 2nd diff = +4
a = 4 ÷ 2 = 2, so sequence contains 2n2
Subtract 2n2 (2, 8, 18, 32) from (4, 11, 22, 37):
Remainder sequence is 2, 3, 4, 5 ⇒ nth term is n + 1
Answer: 2n2 + n + 1

Example 2: A sequence has nth term 3n2 − 2n − 5. Is 128 a term in this sequence?

Solution:

Set 3n2 − 2n − 5 = 128 ⇒ 3n2 − 2n − 133 = 0
Factorise: (3n + 19)(n − 7) = 0
n = 7 or n = −193
Since n = 7 is a positive integer, 128 is the 7th term.
Answer: Yes, it is the 7th term (n = 7)
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Quadratic sequences Questions & Answers

Past Paper Example 1 AQA Nov 2021 1H Q21 4 marks
The first two terms of a quadratic sequence are 10 and 17
Here is some information about the sequence.
1stterm2ndterm3rdterm4thtermSequence1017............Firstdifference+7+13Seconddifference+6+6
Work out an expression for the nth term of the sequence.
Worked Solution & Mark Scheme (4 marks)
Second difference is +6, so a = 62 = 3, giving 3n2 M1
Comparing terms with 3n2:
n = 1: 10 - 3(1)2 = 7 M1
n = 2: 17 - 3(2)2 = 5
n = 3: 30 - 3(3)2 = 3 M1
The sequence of differences 7, 5, 3, … has first difference -2, giving -2n + 9
Therefore, the nth term is 3n2 - 2n + 9. A1
Answer: Answer 3n2 - 2n + 9
Past Paper Example 2 OCR Nov 2022 4H Q18 5 marks
Work out
(b) Here are the first four terms of a sequence.
1   √(3)   3   3√(3)
Write an expression for the nth term.
(c) Here are the first four terms of a quadratic sequence.
-1   5   13   23
The nth term is n2 + bn + c.
Find the value of b and the value of c.
(b) Here are the first four terms of a sequence.
1   √(3)   3   3√(3)
Write an expression for the nth term.
[2 marks]
(c) Here are the first four terms of a quadratic sequence.
-1   5   13   23
The nth term is n2 + bn + c.
Find the value of b and the value of c.
[3 marks]
Worked Solution & Mark Scheme (5 marks)
Part (b) Solution:
First term a = 1 = (√3)⁰, common ratio r = √3. M1
nth term = (√3) {}n-1). A1
Answer (b): (√(3))n-1
Part (c) Solution:
Subtract n² (1, 4, 9, 16) from the sequence (−1, 5, 13, 23): −2, 1, 4, 7. M1
This linear sequence has common difference 3 and nth term 3n − 5. M1
Therefore b = 3 and c = −5. A1
Answer (c): b = 3, c = −5
Past Paper Example 3 GCSE Nov 2024 2H Q15 6 marks
Here are the first five terms of a quadratic sequence.
3     20     47     84     131
(a) Find an expression, in terms of n, for the nth term of this sequence. [3 marks]
(b) find the value of u4 [3 marks]
Worked Solution & Mark Scheme (6 marks)
Part (a) Solution:
1st diffs: 17, 27, 37, 47 M1
2nd diff = 10 ⇒ 5n² M1
Subtracting 5n² gives –2, 0, 2, 4, 6 = 2n – 4 ⇒ 5n² + 2n – 4 A1
Answer (a): 5n2 + 2n - 4
Part (b) Solution:
4 = 9k + k = 10k ⇒ k = 0.4 M1
u₃ = 0.4(4) + 0.4 = 2 M1
u₄ = 0.4(2) + 0.4 = 1.2 A1
Answer (b): u₄ = 1.2

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