Finding nth Terms of Sequences (AQA GCSE Further Maths)
Revision Note
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Linear Sequences
What is a linear sequence?
A linear sequence is one where the terms go up (or down) by the same amount each time
eg 1, 4, 7, 10, 13, … (add 3 to get the next term)
15, 10, 5, 0, -5, … (subtract 5 to get the next term)
A linear sequence is often referred to as an arithmetic sequence
If we look at the differences between the terms, we see that they are constant
What should we be able do with linear sequences?
You should be able to recognise and continue a linear sequence
You should also be able to find a formula for the nth term of a linear sequence in terms of n
This formula will be in the form:
nth term = dn + b, where;
d is the common difference
b is a constant that makes the first term “work”
How do I find the nth term of a linear (arithmetic) sequence?
Find the common difference between the terms – this is d
Put the first term and n = 1 into the formula, then solve to find b
Examiner Tips and Tricks
If a sequence is going up by d each time, then its nth term contains dn
e.g. 5, 7, 9, 11, is going up by 2 each term so the nth term contains 2n
(the complete nth term for this example is 2n + 3)
If a sequence is going down by d each time, then its nth term contains −dn
e.g. 5, 3, 1, -1, ... is going down by 2 each term then the nth term contains −2n
(the complete nth term for this example is −2n + 7)
Worked Example
Given the sequence 5, 7, 9, 11, 13, ...
(a) Find the next three terms.
Looking at the difference between the terms, we see that they are all 2. So this is a linear sequence with common difference 2
So the next three terms are
13 + 2 = 15
15 + 2 = 17
17 + 2 = 19
15, 17, 19
(b) Find a formula for the nth term.
In part (a) we established that the common difference is 2. So d = 2
nth term = 2n + b
The first term is 5. Substitute this and n = 1 into the formula, and solve for b
5 = 2×1 + b
5 = 2 + b
b = 3
Now we can write the nth term
2n + 3
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Quadratic Sequences
What is a quadratic sequence?
Unlike in a linear sequence, in a quadratic sequence the differences between the terms (the first differences) are not constant
However, the differences between the differences (the second differences) are constant
Another way to think about this is that in a quadratic sequence, the sequence of first differences is a linear sequence
eg Sequence: 2, 3, 6, 11, 18, …
1st Differences: 1 3 5 7 (a Linear Sequence)
2nd Differences: 2 2 2 (Constant)
If the second differences there are constant, we know that the example is a quadratic sequence
What should we be able to do with quadratic sequences?
You should be able to recognise and continue a quadratic sequence
You should also be able to find a formula for the nth term of a quadratic sequence in terms of n
This formula will be in the form:
nth term = an2 + bn + c
(The process for finding a, b, and c is given below)
How do I find the nth term of a quadratic sequence?
Work out the sequences of first and second differences
Note: check that the first differences are not constant and the second differences are constant, to make sure you have a quadratic sequence!
e.g. sequence: 1, 10, 23, 40, 61
first difference: 9, 13, 17, 21, ...
second differences: 4, 4, 4, ...
a = [the second difference] ÷ 2
e.g. a = 4 ÷ 2 = 2
Write out the first three or four terms of an2 with the first three or four terms of the given sequence underneath.
Work out the difference between each term of an2 and the corresponding term of the given sequence.e.g. an2 = 2n2 = 2, 8, 18, 32, ...
sequence = 1, 10, 23, 40, ...
difference = -1, 2, 5, 8, ...
Work out the linear nth term of these differences. This is bn + c.
e.g. bn + c = 3n − 4
Add this linear nth term to an2. Now you have an2 + bn + c.
e.g. an2 + bn + c = 2n2 + 3n − 4
Examiner Tips and Tricks
Before doing the very formal process to find the nth term, try comparing the sequence to the square numbers 1, 4, 9, 16, 25, … and see if you can spot the formula
For example:
Sequence 4, 7, 12, 19, 28, …
Square Numbers 1 4 9 16 25
We can see that each term of the sequence is 3 more than the equivalent square number so the formula is
nth term = n2 + 3This could save you a lot of time!
Worked Example
For the sequence 5, 7, 11, 17, 25, ....
(a) Find a formula for the nth term.
Start by finding the first and second differences
Sequence: 5, 7, 11, 17, 25
First differences: 2, 4, 6, 8, ...
Second difference: 2, 2, 2, ...
Hence
a = 2 ÷ 2 = 1
Now write down an2 (just n2 in this case as a = 1) with the sequence underneath, and on the next line write the difference between an2 and the sequence
an2. : 1, 4, 9, 16, ...
sequence: 5, 7, 11, 17, ...
difference: 4, 3, 2, 1, ...
Work out the nth term of these differences to give you bn + c
bn + c = −n + 5
Add an2 and bn + c together to give you the nth term of the sequence
nth term = n2 − n + 5
(b) Hence find the 20th term of the sequence.
Substitute n = 20 into n2 − n + 5
(20)2 − 20 + 5 = 400 − 15
20th term = 385
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