Find a course
Knowledge Hub » GCSEs » GCSE Maths » Sequences using Nth terms

Sequences using Nth terms

What are sequences?

A sequence is just a list of numbers that follows a specific pattern. This can then help us to determine the next number in a sequence by extending the pattern that we have found. With the use of certain rules we can even find the correct value for any number that will occur in the sequence.

A list of numbers such as 1, 5, 9, 13, 17… can be examined and a pattern found. For example, in this sequence we are adding 4 every time and start at the number 1. Different patterns can use different rules such as addition, multiplication, indices etc.

Using the nth term

A very useful technique which we can make use of with sequences is to use the th term to display a sequence which can then be used to find any number in the sequence.

To do this we need to realise that we can label the different values in a list by the times at which they appear. So the first number will be the 1st and then the 2nd and so on. Then we can make a rule to find any number in the list, say the 34th, without having to work out the other 33 numbers that will appear before this value. This will obviously save us a lot of time and effort as we will not need to find irrelevant answers and only the one which we need.


Find a pattern in the numbers 3, 8, 13, 18, 23, 28…

Clearly, the list starts at the number 3 and then each number after increases by 5. Therefore, we can say that each step up the list results in an increase of 5 in the number.

Number in list (n) Result
1 3
2 8
3 13
4 18
5 23
6 28

To determine an equation using the number in the list (n) we must try to find a rule that works for all of the numbers in the above table and for ones that are not incorporated here also.

Since every increase of 1 in the n column results in an increase of 5 in the result we must use the value 5n in our equation to ensure that the jump is correct. However, if this was the equation we would find for say the 3rd result an answer of 5n=5 \times3=15 which is incorrect as the real answer is 13. Therefore, this cannot be our equation.

If we work out the answer that 5n would give us as below then we can find the difference between the values found using this and the actual values that we have from the list given:

Number in list (n) 5n Actual result Difference
1 5 3 -2
2 10 8 -2
3 15 13 -2
4 20 18 -2
5 25 23 -2
6 30 28 -2

From this table we can see that the difference between the actual values and the values for 5n is always -2. Therefore, we can simply add this to our equation and we have an equation that gives the correct answer every time! We end up with 5n-2


  1. First, we need to find the difference between each individual jump in the list of numbers given. This will be the multiplier of in our equation for the sequence.
  2. Now we find the values of the multiplier multiplied by and find the difference between these values and the actual value in the list.
  3. Form an equation using the multiplier of and the difference added on.

Other types of sequences

At times we may come across a sequence which is slightly more complicated than the ones we have just looked at. For your exam you will not be required to calculate rules for a list which involves jumps of different sizes, e.g. 1, 3, 7, 13, 21…

These types of lists are formed from using the number of the value to a certain indices, so they will use n^2n^3 etc. However, since these will not be cropping up on the exam then we will not go into too much detail, but if you wish to research the uses of these types of sequences please feel free.

One type which may come up is the use of a sequence that involves decimals. This would be in the form of \frac{1}{2}\frac{1}{5}\frac{1}{8}\frac{1}{11}

The best way for us to find a sequence that is used for this type of list is to find a sequence for both the numerator and denominator separately. These two equations can then be put as a fraction and will get the correct numerator and denominator for each number in the list. If the list always has a 1 as the numerator then this will obviously be the case in our equation. When it comes to finding the equations for both the top and bottom then we must use the exact same method as before.


Find an equation to find the th term of the following list of numbers:

1, \frac{1}{4}, \frac{1}{7}, \frac{1}{10}, \frac{1}{13}

Firstly, we can see that the initial value of 1 is really the same as \frac{1}{1} so all of the numerators must be 1. Next, we can find the difference between the numbers which are the denominators: this is clearly 3. Therefore, we will have 1 for the numerator and a denominator that uses a multiplier of 3. Putting this in a table to work out if a number should be added or subtracted also gives:

n \frac{1}{3n} Actual value Difference
1 \frac{1}{3} \frac{1}{1} -2 on the denominator
2 \frac{1}{6} \frac{1}{4} -2 on the denominator
3 \frac{1}{9} \frac{1}{7} -2 on the denominator
4 \frac{1}{12} \frac{1}{10} -2 on the denominator
5 \frac{1}{15} \frac{1}{13} -2 on the denominator

So we know that the equation \frac{1}{3n} cannot be correct and the denominator needs a value of -2 adding to it. This will then give us an equation for the nth term of the sequence as:


This same method can always be used to find the equations for a list of fractions.


When we look at sequences we can use functions to show what is done to each number. A function is an operation (such as multiplying, dividing, adding or subtracting) that is done to a number.

For example, we could have a function f(n)=4n+2. This would simply be showing us that we have a function of n (this is the part f(n)) that will multiply n by 4 and then add 2. We can work out the values of the function f(n) in the same way as when we worked out numbers in a sequence.

n f(n) = 4n + 2
1 6
2 10
3 14
4 18
5 22

A function can be anything at all, including the usual operatives you are used to dealing with (adding, dividing, multiplying and dividing) as well as trigonometric functions (sin, cos and tan) or any other type of function. The main thing to remember is that f(n) just means that we have a function of n which means we will be doing something to a group of numbers known as n.

Inverse functions

An inverse function is something that will completely reverse a normal function. This means that, if we take a value and put this into a function as we have just seen, we can then put the answer into another function which will result in the original number. This second function will be the inverse of the first function. An inverse function will be shown with a small indices of minus 1 like this: f^{-1}(n). This shows the inverse of a function f(n).


Find the inverse of the function f(n)=4n+2.

Here we need to find a function f^{-1}(n) which will completely reverse the operations found in the function above. Looking at this logically, we can see that we are multiplying a number by 4 and then adding 2 using the original function. This means that, to completely reverse this process, we need to minus 2 and then divide by 4. So the inverse of f(n) is equal to f^{-1}(n)=\frac{n-2}{4}.

In the example here we have used logic to work backwards. However, there is a more exact way to working out the inverse of a function. To do this we can replace the function with an unknown and then rearrange the equation for n. For example, if we wish to find the inverse to the function f(x)=5x-8 we can replace f(x) with an unknown which would give us:


Rearranging this for the unknown will then give us



We can now replace the original unknown x with the symbol for the inverse function which is f^{-1}(x)=\frac{x+8}{5} and replace the unknown we introduced, a, with x. This leaves us with:


This means that, for any value of x, we could do the function of f(x)=5x-8 followed by the inverse f^{-1}(x)=\frac{x+8}{5} to get back to the original value x. Remember that the unknown value that we have labelled as x here can be any unknown so don’t be confused if this is sometimes an or any other letter.

Composite functions

One way that we can check we have found the correct inverse is to make use of composite functions. This is where we put a number into a combination of different functions. For example, we could find the function f(g(x)) which is the function g(x) within the function f(x). To do this we must start with the function within the brackets and then work outwards. So g(x)=3x+2, and f(x)=\frac{x}{2}. We will be able to work out f(g(4)) by finding g(4) and then putting the answer into f(x). Working through the steps we get:





So we have the final answer of f(g(4))=7

The way in which this is useful for inverse functions is that, if we make a composite with a function’s inverse they will cancel each other out and do nothing to the input number. Therefore, f^{-1}(f(x))=1 for any function. We can test this using the function that we looked at in the previous section which is f(n)=5n-8 and f^{-1}(n)=\frac{n+8}{5}. By finding f^{-1}(f(n)) for the value 3 we get:


This gives us:


So we can see that we have ended up with the same start value of 3.

Composite functions can be built up of any number of individual functions and can vary in difficulty. For the purposes of this course, it is very unlikely that you will ever be asked to work with composite functions of more than two. However, when working with these composite functions you need to follow strict rules in your working:

Working with a composite function

  1. Start with the function that is inside another.
  2. Put the given value into this function and work out the answer.
  3. Take this answer and put it into the other function to give the full answer for the whole composite function.
GCSE Mathematics course

Interested in a Maths GCSE?

We offer the Edexcel IGCSE in Mathematics through our online campus.

Learn more about our maths GCSE courses

Read another one of our posts