 Linear Sequences | Finding the nth Term | MME

# Sequences and Nth Term Worksheets, Questions and Revision

Level 4 Level 5

## What you need to know

### Linear Sequences

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 linear sequences, sometimes called arithmetic progressions. For example, the square numbers

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

There are two different ways you will be expected to work out 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.

Sequences are often denoted by $u_n$, so, in order, the terms of that sequence are

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

This topic focuses on linear sequences and finding the nth term for quadratics, please visit the quadratic sequences revision page. ### Take Note:

There is another type of sequence not mentioned here, known as a geometric progression. 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 geometric progressions. 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. ### Example 1: Linear Sequences – Finding the Terms

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. ### Example 2: Linear Sequences – Nth Term

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. In this example, $a=4$, because $a$ is always the difference between each term. 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

$u_n=4n-1$.

### Example 3: The Fibonacci Sequence

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

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

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

$u_3=1+1=2$

You may have spotted a pattern here. What the algebra is basically saying is we add the previous 2 terms in the sequence to get the next term.

Thus, the first 5 terms in the sequence are: 1, 1, 2, 3, and 5. This is actually the Fibonacci sequence, a famous sequence that you need to recognise. ### Example 4: Linear Sequences – Nth Term

Find the nth term formula for the sequence $-2, 5, 12, 19, 26$.

The first step is to find the difference between each term.  Again the differences are the same, so it is a linear sequence, and thus must have an nth term

$u_n=an+b,$

$a=7$

To work out $b$, consider the sequence formed by putting $n=1, 2, 3, 4, 5$ into $u_n=4n$:

$7, 14, 21, 28, 35$

The difference between these numbers and our sequence is we need to subtract 9 from each term. Thus, our nth term is

$u_n=7n-9$.

### Example Questions

a) To find the $12^{th}$ term of this sequence, we will substitute $n=12$ into the formula given.

$4(12)+1=49$

So, the $12^{th}$ term is 49

b) Every term in this sequence is generated when an integer value of n is substituted into $4n+1$

Hence if we set 77 to equal $4n+1$, we can determine its position in the sequence,

$4n+1=77$

making n the subject by subtracting 1 then dividing by 4,

$n=\dfrac{77-1}{4}=19$

Hence 77 is the $19^{th}$ term in the sequence.

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

\begin{aligned}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 \end{aligned}

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

b) Every term in this sequence is generated when an integer value of n is substituted into $u_n=5n-4$.

If we set 108 to equal $5n-4$, we can determine if it is a part of the sequence or not. If the value of $n$ is a whole number then it is part of the sequence, hence

$5n-4=108$

making n the subject by adding 4 then dividing by 5,

$n=\dfrac{112}{5}=22.4$

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

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 which is 7, hence $a=7$.

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

$u_n=7n-9$

We are told it is an arithmetic progression and so must have an $n^{th}$ formula, $u_n=a+(n-1)d$ where $a$ is the first term in the sequence and $d$ is the difference between each term.

Considering the sequence, $a=2$ and $d=3$

Hence, the $n^{th}$ term formula for our sequence in question is

\begin{aligned}u_n &=2+(n-1)3 \\ &=3n+2-3 \\ &=3n-1 \end{aligned}

We are told it is an arithmetic progression and so must have an $n^{th}$ formula, $u_n=a+(n-1)d$ where $a$ is the first term in the sequence and $d$ is the difference between each term.

Considering the sequence, $a=-213$ and $d=15$

Hence, the $n^{th}$ term formula for our sequence in question is

\begin{aligned}u_n &=-213+(n-1)15 \\ &=15n-213-15 \\ &=15n-228\end{aligned}

Level 4-5

Level 4-5

Level 4-5