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Quadratic sequences Quiz

Grade 8+ • 8F • 10 Questions (37 Marks)

Score: 0 / 10
Quadratic sequences — Notes & Worked Examples Show / Hide

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)
Question 1
[2 marks]
Here are the first six terms of a quadratic sequence.
5 11 21 35 53 75
Find an expression, in terms of n, for the nth term of this sequence.
Worked Solution & Answer
First differences: 6, 10, 14, 18, 22 M1
Second differences: 4 ⇒ 2n²
Subtracting 2n² (2, 8, 18, 32, 50, 72) leaves 3 ⇒ 2n² + 3. A1
Answer: 2n2 + 3
Question 2
[2 marks]
Here are the first five terms of a sequence.
-1   0   3   8   15
Find an expression, in terms of n, for the nth term of this sequence.
Worked Solution & Answer
1st differences: 1, 3, 5, 7 M1
2nd difference = 2, so coefficient of n² is 1
Sequence minus n²: (–1–1, 0–4, 3–9, 8–16, 15–25) = –2, –4, –6, –8, –10 = –2n
nth term = n² – 2n A1
Answer: n2 - 2n
Question 3
[3 marks]
Here are the first four terms of a quadratic sequence.
3 9 17 27
Find an expression, in terms of n, for the nth term of this sequence.
Worked Solution & Answer
1st differences: 6, 8, 10 M2
2nd difference = 2 ⇒ coefficient of n² is 1. Subtracting n² (1, 4, 9, 16) gives 2, 5, 8, 11 = 3n – 1 ⇒ n² + 3n – 1 A1
Answer: n2 + 3n - 1
Question 4
[3 marks]
Here are the first six terms of a quadratic sequence.
–1 5 15 29 47 69
Find an expression, in terms of n, for the nth term of this sequence.
Worked Solution & Answer
1st differences: 6, 10, 14, 18, 22. 2nd differences: 4 M1
⇒ coefficient of n² is 4 ÷ 2 = 2. Sequence minus 2n² (2, 8, 18, 32, 50, 72) gives –3, –3, –3, –3, –3, –3 M1
⇒ 2n² – 3 A1
Answer: 2n2 - 3
Question 5
[3 marks]
Here are the first five terms of a quadratic sequence.
8   19   36   59   88
Find an expression, in terms of n, for the nth term of this sequence.
Worked Solution & Answer
1st diffs: 11, 17, 23, 29 M1
2nd diff = 6 ⇒ 3n² M1
Subtracting 3n² (3, 12, 27, 48, 75) leaves 5, 7, 9, 11, 13 = 2n + 3. Total = 3n² + 2n + 3 A1
Answer: 3n2 + 2n + 3
Question 6 AQA Jun 2025 3H Q18
[3 marks]
Here are the first four terms of a quadratic sequence.
6 15 28 45
Work out an expression for the nth term.
Answer
Worked Solution & Answer
First differences: 9, 13, 17 M1
Second difference = 4 ⇒ 2n² M1
Subtract 2n² (2, 8, 18, 32) from sequence: 4, 7, 10, 13 ⇒ 3n + 1.
nth term = 2n² + 3n + 1. A1
Answer: Answer 2n2 + 3n + 1
Question 7 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.
Answer
Worked Solution & Answer
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
Question 8 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 & Answer
(b)
First term a = 1 = (√3)⁰, common ratio r = √3. M1
nth term = (√3) {}n-1). A1
Answer (b): (√(3))n-1
(c)
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
Question 9
[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]
u₄ =
Worked Solution & Answer
(a)
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
(b)
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
Question 10
[6 marks]
The nth term of a sequence is given by an2 + bn where a and b are integers.
The 2nd term of the sequence is -2
The 4th term of the sequence is 12
(a) Find the 6th term of the sequence. [4 marks]
(b) Find an expression, in terms of n, for the nth term of this sequence. [2 marks]
Worked Solution & Answer
(a)
4a + 2b = –2 and 16a + 4b = 12 M1
⇒ 8a + 4b = –4 M1
⇒ 8a = 16 M1
⇒ a = 2, b = –5. 6th term = 2(6)² – 5(6) = 72 – 30 = 42 A1
Answer (a): 42
(b)
Second difference = 2 ⇒ n² M1
subtracting n² gives –1, –2, –3, –4, –5 ⇒ n² – n A1
Answer (b): n2 - n
Score: 0 / 10 (0 / 37 marks)
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