Quadratic sequences

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GCSE Maths Algebra Sequences

Quadratic Sequences

Here we will learn about quadratic sequences including how to recognise, use and find the nth term of a quadratic sequence.

There are also quadratic sequences worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck

Quadratic sequences are only studied on the higher GCSE Maths pathway.

What are quadratic sequences?

Quadratic sequences are ordered sets of numbers that follow a rule based on the sequence n² = 1, 4, 9, 16, 25,… (the square numbers).

Quadratic sequences always include an n² term.

The difference between each term in a quadratic sequence is not equal, but the second difference between each term in a quadratic sequence is equal.

Quadratic sequences can also be called quadratic algebraic sequences.

Here are two examples of quadratic sequences:
4, 7, 12, 19, 28 requires adding to work out that the second difference is +2
and
1, −4, −15, −32, −55 requires subtracting to work out that the second difference is −6

Example:

What are quadratic sequences?

How to generate terms of quadratic sequences

Substitute the term number that you want to find as n.
Calculate.

Explain how to generate quadratic sequences in 2 steps

Quadratic sequences worksheet

Get your free quadratic sequences worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Quadratic sequences worksheet

Get your free quadratic sequences worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Related lessons on sequences

Quadratic sequences is part of our series of lessons to support revision on sequences. You may find it helpful to start with the main sequences lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Quadratic sequence formula

The quadratic sequence formula is: $an^{2}+bn+c$

Where,

$a, b$ and $c$ are constants (numbers on their own)

$n$ is the term position

We can use the quadratic sequence formula by looking at the general case below:

Let’s use this to work out the $n^{th}$ term of the quadratic sequence,

4, 5, 8, 13, 20, ...

First let’s find the first and second differences of the sequence:

Comparing this to the general case above, we can see that:

The second difference is equal to $2$ so,

The first difference is equal to $1$ so,

The first term is equal to $4$ so,

We can see that $a = 1,\; b = -2$ and $c = 5$

So the nth of this quadratic sequence using the quadratic sequences formula is:

n^{2}-2n+5

Generate terms of quadratic sequences examples

Example 1: finding terms of a sequence

Find the first three terms of the sequence with nth term 2n² + 4n + 1.

Substitute the term number that you want to find as n.

We want to find terms 1, 2 and 3 so substituting each of these into the formula
2n² + 4n + 1 we get,

First term:
n = 1: (2 × 1² ) + (4 × 1) + 1

Second term:
n = 2: (2 × 2² ) + (4 × 2) + 1

Third term:
n = 3: (2 × 3² ) + (4 × 3) + 1

2 Calculate.

Remember to apply BIDMAS

n = 1: (2 × 1² ) + (4 × 1) + 1 = 7

n = 2: (2 × 2² ) + (4 × 2) + 1 = 17

n = 3: (2 × 3² ) + (4 × 3) + 1 = 31

The first term is 7, the next term is 13 and the term after that is 31.

Example 2: finding terms of a sequence

Find the tenth term of the sequence with nth term 5n² − 6n − 3.

Substitute the term number that you want to find as n.

We want to find term 10 so we need to substitute this into the formula

5n² − 6n − 3.

\[(5 \times 10^{2}) − (6 \times 10) − 3\]

Calculate.

\[(5 \times 10^{2}) − (6 \times 10) − 3 = 437\]

The tenth term is 437.

How to find the nth term of quadratic sequences

In order to find the nth term (general term) of a quadratic sequence we have to find the second difference. To do this, we calculate the first difference between each term in a quadratic sequence and then calculate the difference between this new sequence.

1 Find the first difference (d₁) and second difference (d₂) for the sequence.

2 Halve the second difference

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

3 Subtract an² from the original sequence.

4 If this produces a linear sequence, find the nth term of it.

5 Add the nth term for the linear sequence to an² to work out the nth term of the quadratic sequence.

What are the 5 steps to find the nth term of a sequence?

Step 4 is an additional step dependent on whether there is a remainder in step 3.

If there is no remainder after subtracting an² from the original sequence, step 4 is not needed.
If the same constant remains for each term after the subtraction, we just add it to an².
If a sequence is generated after step 3, continue with step 4.

See the table below.

	1	2	3
Sequence	2, 8, 18, 32, 50	3, 9, 19, 33, 51	5, 12, 23, 38, 57
an²	2, 8, 18, 32, 50	2, 8, 18, 32, 50	2, 8, 18, 32, 50
Step 3	0, 0, 0, 0, 0, …	1, 1, 1, 1, 1, …	3, 4, 5, 6, 7, …
Step 4			n + 2
Step 5	2n²	2n² + 1	2n² + 1 + 2

Here,

Sequence 1 had no remainder and so the sequence is 2n²
Sequence 2 had a remainder of 1 and so this sequence is 2n² + 1
Sequence 3 had another sequence as the remainder and so the nth term of this linear sequence was calculated and added to 2n² to get 2n² + n + 2.

Find the nth term of quadratic sequences examples

Example 3: find the nth term of a quadratic sequence of the form an².

Find the nth term of this quadratic sequence:

2, 8, 18, 32, 50, …

Find the first difference $(d_{1})$ and second difference $(d_{2})$ for the sequence.

Here, the second difference d₂ = 4.

Halve the second difference

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

d₂ = 4,

\[4 \div 2 = 2\]

This means a = 2 and so we have the sequence 2n².

Subtract $an^{2}$ from the original sequence.

Original Sequence	2	8	18	32	50
n²	1	4	9	16	25
2n²	2	8	18	32	50
OS − 2n²	0	0	0	0	0

Here, the remainder for each term is 0. This means that we have found the nth term of the quadratic sequence.

Find the $n$ th term for the linear sequence generated.

Not required for this example as the remainder is 0 for each term.

Add the $n$ th term for the linear sequence to $an^{2}$ to find the $n$ th term of the quadratic sequence.

The nth term of the quadratic sequence is 2n².

Example 4: find the nth term of a quadratic sequence of the form an² + c.

Find the nth term rule of the quadratic sequence:

5, 14, 29, 50, 77, …

Find the first difference $(d_{1})$ and second difference $(d_{2})$ for the sequence.

Here, the second difference d₂ = 6.

Halve the second difference.

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

d₂ = 6,

\[6 \div 2 = 3\]

This means a = 3 and so we have the sequence 3n².

Subtract $an^{2}$ from the original sequence.

Original Sequence	5	14	29	50	77
n²	1	4	9	16	25
3n²	3	12	27	48	75
OS − 3n²	2	2	2	2	2

Here, the remainder for each term is 2.

Find the $n$ th term for the linear sequence generated.

Not required for this example as the remainder is 2 for each term.

Add the $n$ th term for the linear sequence to $an^{2}$ to find the $n$ th term of the quadratic sequence.

The nth term of the quadratic sequence is 3n² + 2.

Example 5: find the nth term of a quadratic sequence of the form an² + bn + c.

Find the nth term of this quadratic sequence:

9, 18, 31, 48, 69, …

Find the first difference $(d_{1})$ and second difference $(d_{2})$ for the sequence

Here, the second difference d₂ = 4.

Halve the second difference

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

d₂ = 4,

\[4 \div 2 = 2\]

This means a = 2 and so we have the sequence 2n².

Subtract $an^{2}$ from the original sequence.

Original Sequence	9	18	31	48	69
n²	1	4	9	16	25
2n²	2	8	18	32	50
OS − 2n²	7	10	13	16	19

Here, the remainder generates a linear sequence and so we must find the nth term for this sequence as well.

Find the $n$ th term for the linear sequence generated.

Here, the nth term for the remainder is 3n + 4.

Add the $n$ th term for the linear sequence to $an^{2}$ to find the $n$ th term of the quadratic sequence.

Here, the nth term of the quadratic sequence is 2n² + 3n + 4.

Example 6: find the nth term of a quadratic sequence of the form an² + bn + c.

Find the nth term of this quadratic sequence:

19, 36, 61, 94, 135, …

Find the first difference $(d_{1})$ and second difference $(d_{2})$ for the sequence.

Here, the second difference d₂ = 8.

Halve the second difference.

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

d₂ = 8,

\[8 \div 2 = 4\]

This means a = 4 and so we have the sequence 4n².

Subtract $an^{2}$ from the original sequence.

Original Sequence	19	36	61	94	135
n²	1	4	9	16	25
4n²	4	16	36	64	100
OS − 4n²	15	20	25	30	35

Here, the remainder generates a linear sequence and so we must find the $n$ th term for this sequence as well.

Find the $n$ th term for the linear sequence generated.

Here, the nth term for the remainder is 5n + 10.

Add the $n$ th term for the linear sequence to $an^{2}$ to find the $n$ th term of the quadratic sequence.

Here, the nth term of the quadratic sequence is 4n² + 5n + 10.

Example 7: find the nth term of a quadratic sequence of the form an² + bn + c where terms are generated using a pattern.

In the diagram below, the area of each pool can be represented by a quadratic sequence.

Find the area for a pool of width n.

Find the first difference $(d_{1})$ and second difference $(d_{2})$ for the sequence.

Here, the second difference d₂ = 2.

Halve the second difference.

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

d₂ = 2,

\[2 \div 2 = 1\]

This means a = 1 and so we have the sequence n².

Subtract $an^{2}$ from the original sequence.

Original Sequence	1	4	9	16	25
n²	1	4	9	16	25
OS − n²	0	0	0	0	0

Here, the remainder for each term is 0. This means that we have found the nth term of the quadratic sequence.

Find the $n$ th term for the linear sequence generated.

Not required for this example as the remainder is 0 for each term.

Add the $n$ th term for the linear sequence to $an^{2}$ to find the $n$ th term of the quadratic sequence.

The nth term of the quadratic sequence is n².

So the area of the pool with width n would be n².

Example 8: find the nth term of a quadratic sequence of the form an² + bn + c where a < 0.

Find the nth term of this quadratic sequence:

8, -10, -34, -64, -100, …

Find the first difference $(d_{1})$ and second difference $(d_{2})$ for the sequence.

Here, the second difference d₂ = −6.

Halve the second difference.

\[\left ( \frac{d_{2}}{2} =a \right )\]

to find a, the coefficient of n².

d₂ = −6,

\[−6 \div 2 = −3\]

This means a = −3 and so we have the sequence −3n².

Subtract $an^{2}$ from the original sequence.

Original Sequence	8	-10	-34	-64	-100
n²	1	4	9	16	25
−3n²	-3	-12	-27	-48	-75
OS − (−3n² )	11	2	-7	-16	-25

Here, the remainder generates a linear sequence and so we must find the nth term for this sequence as well.

Find the $n$ th term for the linear sequence generated.

Add the $n$ th term for the linear sequence to $an^{2}$ to find the nth term of the quadratic sequence.

Here, the nth term of the quadratic sequence is −3n² − 9n + 20.

Common misconceptions

The value of the second difference is not halved before using it as the coefficient of n²

For example, for the quadratic sequence n² = 1, 4, 9, 16, 25, … the second difference is equal to 2 and so the nth term is incorrectly written as 2n².

The quadratic sequence is answered as if it were an arithmetic sequence

For a quadratic sequence we will have a common second difference.

When the sequence is decreasing into negative terms, the second difference is written incorrectly as a positive number

For example, the sequence −n² = −1, −4, −9, −16, … has a second difference of -2 but is written as 2.

The remainder is subtracted from the an² term rather than added as it was found by subtraction

For example, the remainder for the sequence n(n + 2) is the sequence 2n but is incorrectly subtracted to get the nth term of n(n − 2).

The value for n cannot be negative

n represents the term (position) numbers and therefore it can only be positive integers starting from 1 and should also not include 0, n = 1, 2, 3, 4, 5, ….

This is important when finding the term in the sequence given its value as a zero or negative solution for n can be calculated.

The value for the term is used instead of n

For example, find n when the term has a value of 25 means that the nth term = 25 and not n = 25.

Practice quadratic sequences questions

-7, -6, -3, 4, 15

-5, -3, -1, 1, 3

-5, -6, -3, 4, 15

9, 22, 39, 60, 85

2 \times 1^{2} \hspace{1mm}- 7 \times 1 = -5\\ 2 \times 2^{2} \hspace{1mm}- 7 \times 2 = -6\\ 2 \times 3^{2} \hspace{1mm}- 7 \times 3 = -3\\ 2 \times 4^{2} \hspace{1mm}- 7 \times 4 = 4\\ 2 \times 5^{2} \hspace{1mm}- 7 \times 5 = 15\\

3n+2

n^{2}-3

2n^{2}-3n^{2}

n^{2}+3

The second difference is $2$

2 \div 2 = 1

Therefore it is $n^{2}$

\begin{aligned} &Original \; Sequence \quad \quad -2 \quad \quad 1 \quad \quad 6 \quad \quad 13 \quad \quad 22\\ &n^2 \hspace{3.7cm} 1 \quad \quad 4 \quad \quad 9 \quad \quad 16 \quad \quad 25\\ &OS-n^2 \hspace{2.5cm} -3 \quad -3 \quad -3 \quad \;-3 \quad \;\;-3 \end{aligned}

The remainder is $4$ for each term therefore our sequence is

n^{2}-3

4n^{2}+n-5

4n^{2}-5n

8n^{2}-8n

n^{2}+4n-5

The second difference is $8.$

8 \div 2 = 4

Therefore it is $4n^{2}$

\begin{aligned} &Original \; Sequence \quad \quad 0 \quad \quad 13 \quad \quad 34 \quad \quad 63 \quad \quad 100\\ &n^2 \hspace{3.2cm} \;1 \quad \quad\; 4 \quad \quad \;\;9 \quad \quad 16 \quad \quad\; \;25\\ &4n^2 \hspace{3.1cm} 4 \quad \quad 16 \quad \quad 36 \quad \quad 64 \quad \quad 100\\ &OS-n^2 \hspace{2cm}\; -4 \quad \;-3 \quad \;\; -2 \quad \;\;-1 \quad \quad \;\;0 \end{aligned}

The nth term of $-4, -3,-2, -1, 0, \dots$ is $n-5$ so our sequence is

4n^{2}+n-5

0.1n^{2}+2n

10n^{2}+2n

2n^{2}-10

0.1n^{2}+2

The second difference is $0.2$

0.2 \div 2 = 0.1

Therefore it is $0.1n^{2}$

\begin{aligned} &Original \; Sequence \quad \quad 2.1 \quad \quad 4.4 \quad \quad 6.9 \quad \quad 8.6 \quad \quad 10.5\\ &n^2 \hspace{3.4cm} 1 \quad \quad\; \;\;4 \quad \quad \;\;\;9 \quad \quad \;16 \quad \quad\; \;25\\ &0.1n^2 \hspace{2.9cm} 0.1 \quad \quad 0.4 \quad \quad 0.9 \quad \;\;\;1.6 \quad \quad \;2.5\\ &OS-0.1n^2 \hspace{2.1cm} 2 \quad\quad\;\; \;4 \quad \quad \;\;\;6 \quad \quad \;\; 8 \quad \quad \;\;\;10 \end{aligned}

The nth term of $2, 4, 6, 8, 10$ is $2n$ so our sequence is

0.1n^{2} +2n

n^{2}-4

4n^{2}

n^{2}+4

2n^{2}+3

Number of tiles: $5, 8, 13, 20$

The second difference is $2$

2 \div 2 = 1

Therefore it is $n^{2}$

\begin{aligned} &Original \; Sequence \quad \quad 5 \quad \quad 8 \quad \quad 13 \quad \quad 20 \quad \quad 29\\ &n^2 \hspace{3.2cm} \;1 \quad \quad 4 \quad \quad \;9 \quad \;\quad 16 \quad \quad 25\\ &OS-n^2 \hspace{2.4cm} 4 \quad \quad 4 \quad \quad \; 4 \quad \quad \;\;4 \quad \quad \;\;4 \end{aligned}

The remainder is 4 each time so the nth term is

n^{2}+4

a=-1,\; b=6

a=7,\; b=-2

a=4,\; b=1

a=6,\; b=-1

The second difference is $-2$

-2 \div 2 = -1

Therefore it is $-n^{2}$

\begin{aligned} &Original \; Sequence \quad \quad 5 \quad \quad 2 \quad \; -3 \quad \; -10 \quad \; -19\\ &-n^2 \hspace{2.5cm} \;-1 \quad -4 \quad \; -9 \quad \;-16 \quad \; -25\\ &OS-n^2 \hspace{2.4cm} 6 \quad \quad 6 \quad \quad \; 6 \quad \quad \;\;6 \quad \quad \;\;6 \end{aligned}

The nth term is $-n^2+6,$ which can also be written as $6-n^2.$

$a=6$ and $b=-1.$

Quadratic sequences GCSE questions

1. Work out the first four terms in the sequence with $n$ th term $2n^{2} + 5n – 3$

(2 marks)

Show answer

\begin{aligned} 2 \times 1^{2} + 5 \times 1  –  3 &=4 \\ 2 \times 2^{2} + 5 \times 2  –  3 &=15\\ 2 \times 3^{2} + 5 \times 3  –  3 &=30\\ 2 \times 4^{2} + 5 \times 4  –  3 &=49 \end{aligned}

$= 4, 15, 30, 49$

(2)

(a) Here is a patterned sequence made from equilateral triangles.

Find the $n$ th term formula for the number of triangles used to form each pattern.

(b) How many triangles would be used in the 20th pattern?

(6 marks)

Show answer

(a)

Sequence is $1, 3, 6, 10$

First differences: $2, 3, 4$

Second differences: $1, 1$

(1)

$0.5n^{2}$

(1)

Original Sequence	1	3	6	10
n²	1	4	9	16
0.5n²	0.5	2	4.5	8
OS − (0.5n² )	0.5	1	1.5	2

(1)

$N$ th term of $0.5, 1, 1.5, 2$ is $0.5n$ so formula is

$0.5n^{2} + 0.5n$

(1)

(b)

$0.5 \times 20^{2} + 0.5 \times 20$

(1)

$210$

(1)

3. The first four terms in a sequence are:

$12, 20, 30, 42, 56$

(a) Find the $n$ th term formula for this sequence.

(b) Hence find which term has value $110$ .

(7 marks)

Show answer

(a)

First differences: $8, 10, 12, 14$

Second differences: $2, 2, 2$

(1)

$n^{2}$

(1)

Original Sequence	12	20	30	42	56
n²	1	4	9	16	25
OS-0n²	11	16	21	26	31

(1)

$N$ th term of $11, 16, 21, 26, 31$ is $5n+6$ so formula is

$n^{2}+5n+6$

(1)

(b)

$n^{2}+5n+6=110$

(1)

$\begin{aligned} n^{2}+5n-104&=0 \\ (n-8)(n+13)&=0 \end{aligned}$

(1)

$n = 8, 8th$ term

(cannot have $-13th$ term)

(1)

Learning checklist

You have now learned how to:

recognise and use quadratic sequences

deduce expressions to calculate the nth term of quadratic sequences.

The next lessons are

Still stuck?

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