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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.

**Quadratic sequences** are ordered sets of numbers that follow a rule based on the sequence n^{2} = 1, 4, 9, 16, 25,… (the square numbers).

So quadratic sequences always include an n^{2} 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

E.g.

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

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

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

COMING SOONTo find out more about the different types of sequences, and how to answer sequence related questions you may find it helpful to look at the introduction to sequences lesson or one of the others in this section.

- Sequences
- Arithmetic sequence
- Nth term
- Quadratic nth term
- Geometric sequence

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+5Find the first three terms of the sequence with nth term ^{2} + 4n + 1.

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

We want to find terms ^{2} + 4n + 1

First term:

n=1: (^{2} ) + (4 × 1) + 1

Second term:

n=2: (^{2} ) + (4 × 2) + 1

Third term:

n=3: (^{2} ) + (4 × 3) + 1

2 Calculate.

Remember to apply BIDMAS

n=1: (2 × 1^{2} ) + (4 × 1) + 1

n=2: (2 × 2^{2} ) + (4 × 2) + 1

n=3: (2 × 3^{2} ) + (4 × 3) + 1

The first term is

Find the tenth term of the sequence with ^{2} − 6n − 3

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

We want to find term

^{2} − 6n − 3.

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

Calculate.

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

The tenth term is

In order to find the

1 Find the first difference _{1})_{2})

2 Halve the second difference

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

to find ^{2}

3 Subtract ^{2}

4 If this produces a linear sequence, find the

5 Add the nth term for the linear sequence to ^{2}

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

- If there is no remainder after subtracting
an from the original sequence, step 4 is not needed.^{2} - If the same constant remains for each term after the subtraction, we just add it to
an .^{2} - 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 |

an^{2} | 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 | 2n^{2} | 2n^{2} + 1 | 2n^{2} + 1 + 2 |

Here,

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

Find the

Find the first difference (*d*_{1}) and second difference (*d*_{2}) for the sequence.

Here, the second difference _{2} = 4

Halve the second difference

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

to find ^{2}

_{2} = 4,

\[4 \div 2 = 2\]

This means ^{2}

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

Original Sequence | 2 | 8 | 18 | 32 | 50 |

n^{2} | 1 | 4 | 9 | 16 | 25 |

2n^{2} | 2 | 8 | 18 | 32 | 50 |

OS − 2n^{2} | 0 | 0 | 0 | 0 | 0 |

Here, the remainder for each term is

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

Not required for this example as the remainder is

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

The nth term of the quadratic sequence is ^{2}

Find the

Find the first difference (*d*_{1}) and second difference (*d*_{2}) for the sequence.

Here, the second difference _{2} = 6

Halve the second difference.

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

to find ^{2}

_{2} = 6,

\[6 \div 2 = 3\]

This means ^{2}

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

Original Sequence | 5 | 14 | 29 | 50 | 77 |

n^{2} | 1 | 4 | 9 | 16 | 25 |

3n^{2} | 3 | 12 | 27 | 48 | 75 |

OS − 3n^{2} | 2 | 2 | 2 | 2 | 2 |

Here, the remainder for each term is

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

Not required for this example as the remainder is

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

The nth term of the quadratic sequence is ^{2} + 2.

Find the nth term of this quadratic sequence:

Find the first difference (*d*_{1}) and second difference (*d*_{2}) for the sequence

Here, the second difference _{2} = 4

Halve the second difference

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

to find ^{2}

_{2} = 4,

\[4 \div 2 = 2\]

This means ^{2}

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

Original Sequence | 9 | 18 | 31 | 48 | 69 |

n^{2} | 1 | 4 | 9 | 16 | 25 |

2n^{2} | 2 | 8 | 18 | 32 | 50 |

OS − 2n^{2} | 7 | 10 | 13 | 16 | 19 |

Here, the remainder generates a linear sequence and so we must find the

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

Here, the nth term for the remainder is

Add the nth 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 ^{2} + 3n + 4

Find the nth term of this quadratic sequence:

Find the first difference (*d*_{1}) and second difference (*d*_{2}) for the sequence.

Here, the second difference _{2} = 8

Halve the second difference.

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

to find ^{2}

_{2} = 8,

\[8 \div 2 = 4\]

This means ^{2}

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

Original Sequence | 19 | 36 | 61 | 94 | 135 |

n^{2} | 1 | 4 | 9 | 16 | 25 |

4n^{2} | 4 | 16 | 36 | 64 | 100 |

OS − 4n^{2} | 15 | 20 | 25 | 30 | 35 |

Here, the remainder generates a linear sequence and so we must find the

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

Here, the nth term for the remainder is

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

Here, the ^{2} + 5n + 10.

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

Find the area for a pool of width

Find the first difference (*d*_{1}) and second difference (*d*_{2}) for the sequence.

Here, the second difference _{2} = 2

Halve the second difference.

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

to find ^{2}

_{2} = 2,

\[2 \div 2 = 1\]

This means ^{2}

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

Original Sequence | 1 | 4 | 9 | 16 | 25 |

n^{2} | 1 | 4 | 9 | 16 | 25 |

OS − n^{2} | 0 | 0 | 0 | 0 | 0 |

Here, the remainder for each term is

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

Not required for this example as the remainder is

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

The ^{2}

So the area of the pool with width ^{2}

Find the nth term of this quadratic sequence:

Find the first difference (*d*_{1}) and second difference (*d*_{2}) for the sequence.

Here, the second difference _{2} = −6

Halve the second difference.

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

to find ^{2}

_{2} = −6,

\[−6 \div 2 = −3\]

This means ^{2}

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

Original Sequence | 8 | -10 | -34 | -64 | -100 |

n^{2} | 1 | 4 | 9 | 16 | 25 |

−3n^{2} | -3 | -12 | -27 | -48 | -75 |

OS − (−3n^{2} ) | 11 | 2 | -7 | -16 | -25 |

Here, the remainder generates a linear sequence and so we must find the

Find the nth term for the linear sequence generated.

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

Here, the ^{2} − 9n + 20.

**The value of the second difference is not halved before using it as the coefficient of**n ^{2}

For example, for the quadratic sequence ^{2} = 1, 4, 9, 16, 25^{2}

**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 ^{2} = −1, −4, −9, −16,

**The remainder is subtracted from the**an term rather than added as it was found by subtraction^{2}

For example, the remainder for the sequence

**The value for**n cannot be negative

This is important when finding the term in the sequence given its value as a zero or negative solution for

**The value for the term is used instead of**n

For example, find

1. Work out the first five terms of the sequence 2n^2-7n

-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\\

2. Find the nth term of the quadratic sequence:

-2, 1, 6, 13, 22

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

3. The first five terms of a quadratic sequence are 0, 13, 34, 63 and 100 . By calculating the second difference, find the nth term of the sequence.

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, … is n-5 so our sequence is

4n^{2}+n-5

4. Find the nth term of the decimal quadratic sequence:

2.1, 4.4, 6.9, 8.6, 10.5

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

5. Below are the first four terms of a patterned sequence. Calculate the formula to work out the number of tiles in pattern n.

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

6. The first five terms of a quadratic sequence are 5, 2, -3, -10, and -19 . The formula for the nth term of this sequence can be written in the form a+bn^{2} where a and b are integers. Find the values of a and b .

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.

1. Work out the first four terms in the sequence with n th term 2 \times n^{2} + 5 \times n – 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)**

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 |

n^{2} |
1 | 4 | 9 | 16 |

0.5n^{2} |
0.5 | 2 | 4.5 | 8 |

OS − (0.5n^{2} ) |
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 |

n^{2} |
1 | 4 | 9 | 16 | 25 |

OS-0n^{2} |
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)**

You have now learned how to:

- recognise and use quadratic sequences
- deduce expressions to calculate the nth term of quadratic sequences.

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