Linear (Arithmetic) Sequences Revision Notes for Leaving Cert Mathematics

Linear (Arithmetic) Sequences

Introduction

infoNote

A sequence is a set of objects (usually numbers) listed in order.

infoNote

An arithmetic sequence is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant difference is called the common difference, denoted as $d$ .

The first term of any arithmetic is denoted as $a$ , so in general, an arithmetic sequence is :

a,a+d,a+2n, a+3d,a+4d, ...

Each positional terms is denoted as $T_{n}$ where $n$ is the position of the term. So $T_1=a, T_2=a+d$ and so on.

The general term of any arithmetic sequence is

infoNote

General term of an arithmetic sequence : Page 22

T_n=a+(n-1)d

Example

infoNote

List the first $5$ terms of arithmetic sequence with $a=6$ and $d=-2$ .

$T_1=a+(1-1)(-2)=a=6$
$T_2=6+(2-1)(-2)=4$
$T_3=6+(3-1)(-2)=2$
$T_4=6+(4-1)(-2)=0$
$T_5=6+(5-1)(-2)=-2$

Example

infoNote

If $6,p, q,32$ are the first four consecutive terms in an arithmetic sequence, find $p$ and $q$ , and then find the general term.

$T_1=a+(1-1)(d)=a=6$
$T_4=6+(4-1)(d)=32$

\begin{align*} T_4=6+3d&=32 \\ 3d&=26 \\ d&=\tfrac{26}{3} \end{align*}

Once $a$ and $d$ are known, we can find any term, $T_n$ .

$p=T_2=(6+\tfrac{26}{3})=\tfrac{44}{3}$
$q=T_3=(\tfrac{44}{3}+\tfrac{26}{3})=\tfrac{70}{3}$ The general term :
$T_n=a+(n-1)d=6+(n-1)\tfrac{26}{3}$

Example

infoNote

Which term is $249$ in the arithmetic sequence $2,15,28,...$ ?

First let's determine the common difference, which can be found by taking any term subtracted by the term before it, in general :

d=T_n-T_{n-1}

$T_3-T_2=28-15=13=d$

Now we are looking for the position of the term $249$ , so $n$ is the unknown variable :

Let $T_n=249$ . Use the general term formula :

\begin{align*} T_n&=a+(n-1)d \\ 249&=2+(n-1)(13) \\ 247&=13n-13 \\ 260&=13n \\ 20&=n \end{align*}

$n$ must be a natural number, so getting anything that isn't a natural number indicates a mistake.

:::

Proving a Sequence is Arithmetic

To show that any sequence is arithmetic, we need to prove that the common difference is constant, in general, if a sequence is arithmetic then :

d=T_n-T_{n-1}=c

where $c$ is some constant.

Example

infoNote

Determine if the general term $T_n=n^2+3n$ forms an arithmetic sequence.

We need $T_n$ (given to us already) and $T_{n-1}$ , so we can just sub in $n-1$ for $n$ .

\begin{align*} T_n&=n^2+3n \\ T_{n-1}&=(n-1)^2+3(n-1) \\ &=n^2-2n+1+3n-3 \\ &=n^2+n-2 \end{align*}

Now determine if $T_n-T_{n-1}$ is constant :

\begin{align*} T_n-T_{n-1}&=(n^2+3n)-(n^2+n-2) \\ &=n^2+3n-n^2-n+2 \\ &= 2n+2 \end{align*}

$2n+2$ is not constant, so the sequence formed from $T_n$ is not arithmetic.

Linear (Arithmetic) Sequences (Leaving Cert Mathematics): Revision Notes