## What you need to know

In maths, a sequence is a list of numbers, algebraic terms, shapes, or other mathematical objects. Most of the sequences you’ll come across will be number sequences. The sequences that we’re interested in are the ones that follow a pattern or a rule. For example, the square numbers

$1, 4, 9, 16, 25, 36, ...$

form a sequence. This is one sequence you would be expected to recognise. Others include the cube numbers: $1, 8, 27, 64, 125$, the triangular numbers: $1, 3, 6, 10, 15, ...$, and the rest of the kinds of sequences we’ll see in this topic.

There are two different ways you will be expected to generate (work out the terms in) a sequence:

• A term-to-term rule – each term in the sequence is calculated by performing a fixed set of operations (such as “multiply by 2 and add 3”) to the term(s) before it.
• A position-to-term rule – each term in the sequence is calculated according to its position in the sequence. Usually, this takes the form of an nth term formula, which is a formula that takes in the position of a term (for the first term, $n=1$, and so on) and outputs the value of that term.

There’s some notation that needs mentioning. Sequences are often denoted by $u_n$, so, in order, the terms of that sequence are

$u_1, u_2, u_3, u_4, ...$

And so on. We’ll see how this notation can be used further in the next couple of examples.

Example: Generate the first 4 terms of the sequence defined by the nth term formula

$u_{n}=3n-2$

All we need to do is input the first 4 values of $n: 1, 2, 3,$ and $4$ into the formula given to get the first 4 terms of the sequences. So, we get

$u_1=3(1)-2=1,\,\,\,\,\,\,\,\,\,\,\,u_2=3(2)-2=4$

$u_3=3(3)-2=7,\,\,\,\,\,\,\,\,\,\,\,u_4=3(4)-2=10$

So, the first 4 terms of this sequence are 1, 4, 7, and 10. This type of sequence is known as a linear sequence (or, occasionally, an arithmetic progression), and is recognisable by the fact that all linear sequences have the same difference between each term, which in this case is 3. More on this soon.

Example: Generate the first 5 terms of the sequence defined by the term-to-term formula

$u_{n+1}=u_{n}+u_{n-1}\,\,\,\,\text{ with }\,\,\,\,u_1=1\text{ and }u_2=1.$

Okay, so there’s a bit more going on here, but it’s not as bad as it seems. If we choose $n$ to be 2, then $n+1=3$ and $n-1=1$, so the formula looks like

$u_3=u_2+u_1$

We’re given the values of $u_1$ and $u_2$ in the question, so we get that

$u_3=1+1=2$

Here’s what you need to understand to make sense of this type of sequence: if $n$ is some fixed number, then $n+1$ must be the number after it, and $n-1$ must be the number before it. Therefore, we can infer that $u_{n+1}$ is the term after $u_n$, which in turn is the term after $u_{n-1}$. Thus, the formula

$u_{n+1}=u_{n}+u_{n-1}$

is saying: to get the next term in this sequence, you need to add together the two terms before it. So, the two remaining terms we need from this sequence are

$u_4=u_3+u_2=2+1=3\,\,\,\text{ and }\,\,\,u_5=u_4+u_3=3+2=5$

Thus, the first 5 terms in the sequence are: 1, 1, 2, 3, and 5. This is actually the Fibonacci sequence, a famous sequence (to mathematicians, at least) that you should be able to recognise. You may see sequences that have a similar idea but are slightly different, e.g. they have different starting values.

Now, in the first example we saw how to generate a sequence from a nth term formula. Now, we’re going to look at finding an nth term formula of a linear sequence when given the first few terms.

Example: Find the nth term formula for the sequence $3, 7, 11, 15, 19$.

The first step is to find the difference between each term.

The differences are the same, so it must be a linear sequence, and thus must have nth term formula

$u_n=an+b,$

Where $a$ and $b$ are numbers to be determined. Fortunately, we’ve already figured out one: $a=4$, because $a$ is always the difference between each term. So, the nth term formula is $u_n=4n+b$. To work out $b$, consider the sequence formed by putting $n=1, 2, 3, 4, 5$ into $u_n=4n$:

$4, 8, 12, 16, 20$

What’s the difference between these terms and our actual sequence? They’re all too big by 1. So, to make our original sequence, we must subtract 1 from $4n$. Thus, our nth term formula is

$u_n=4n-1$.

There is another type of sequence not mentioned here, known as a geometric progression. In a linear sequence, you add a fixed value to each term to get to the next. In a geometric progression, you multiply each term by a fixed value to get to the next. They are defined by the nth term formula

$u_n=r^n$

Where $r$ is usually a whole number or fraction. For example, the sequence defined by $u_n=3^n$ is

$3, 9, 27, 81, 243, ...$

You should be able to recognise sequences of this type. If you are on the higher course, you may come across geometric progression where $r$ is a surd. As you can see, sequences come in all shapes and sizes. It’s worth seeing as many different kinds as you can to be best prepared for whatever you come up against.

## Example Questions

a) To generate the first 5 terms of this sequence, we will substitute $n=1, 2, 3, 4, 5$ into the formula given.

$u_1=5(1)-4=1$

$u_2=5(2)-4=6$

$u_3=5(3)-4=11$

$u_4=5(4)-4=16$

$u_5=5(5)-4=21$

So, the first 5 terms are: 1, 6, 11, 16, and 21.

b) Every term in this sequence must be some output of the formula $u_n=5n-4$. So, if we make a term in the sequence equal to $5n-4$, then we can rearrange it to find the value of $n$ that generates that term (aka the position of that term in the sequence). If we do this with 108, then

$5n-4=108$

Add 4 to both sides to get

$5n=112$

Then, if we divide both sides by 5, we see that

$n=112\div5=22.4$

This is not a whole number. Since there is no “22.4th” position in the sequence, it must the case that 108 is not a term in this sequence.

So, firstly we know that if we add the first two terms that would give us the third. As an equation:

$a+b=8$

This doesn’t tell us much, yet. However, we also know that if we add the 2nd and 3rd term, we get the 4th. Writing this as an equation, we get

$b+8=11$

Then, subtracting 8 from both sides we get that $b=3$. Now we know the value of $b$, we can put it back into the first equation – the one we got from recognising that the 3rd term is the sum of the first two terms. Doing so, we get

$a+3=8$

Then, subtracting 3 from both sides, we get that $a=5$, and so we are done.

We are told it is an arithmetic progression and so must have nth formula: $u_n=an+b$. To find $a$, we must inspect the difference between each term.

The difference is consistently 7, so $a=7$. Then, to find $b$, let’s consider the sequence generated by $u_n=7n$:

$7,\,\,14,\,\,21,\,\,28,\,\,35$

Every term is this sequence is bigger than the corresponding terms in the original sequence by 9. So, to get to the original sequence, we will have to subtract 9 from every term in this sequence. In other words, the nth term formula for our sequence in question is

$u_n=7n-9$

## Sequences (Linear) Revision and Worksheets

Sequences
Level 4-5
Sequences (2)
Level 4-5
Sequences & nth term
Level 4-5
Sequences: nth Term (2)
Level 4-5
Sequences: find the next terms
Level 4-5
Sequences: Use Position to Term Formula
Level 4-5

## Sequences (Linear) Teaching Resources

Linear sequences and finding the nth term are topics that students often make silly errors on. These errors can certainly be reduced by getting them to practise different questions and exam style problems. At Maths Made Easy we have made the effort to source what we think are the best GCSE Maths linear sequence revision resources. We hope tutors, teachers and students find these useful.