# Arithmetic Sequences

An arithmetic sequence is a sequence of numbers which increases or decreases by a constant amount each term.

We can write a formula for the ${n}^{\text{th}}$ term of an arithmetic sequence in the form

${a}_{n}=dn+c$ ,

where $d$ is the common difference . Once you know the common difference, you can find the value of $c$ by plugging in $1$ for $n$ and the first term in the sequence for ${a}_{1}$ .

Example 1:

$\left\{1,5,9,13,17,21,25,...\right\}$

is an arithmetic sequence with common difference of $4$ .

(Since

$5-1=4$ ,

$9-5=4$ ,

etc.)

To find the next $3$ terms, we just keep adding $4$ :

$\begin{array}{l}25+4=29\\ 29+4=33\\ 33+4=37\end{array}$

So, the next $3$ terms are $29$ , $33$ , and $37$ .

To find a formula for the ${n}^{\text{th}}$ term, we substitute $n=1,{a}_{1}=1$ and $d=4$ in

${a}_{n}=dn+c$

to find $c$ .

$\begin{array}{l}1=4\left(1\right)+c\\ c=-3\end{array}$

So, a formula for the ${n}^{\text{th}}$ term of the sequence is

${a}_{n}=4n-3$ .

Example 2:

$\left\{12,9,6,3,0,-3,-6,...\right\}$

is an arithmetic sequence with common difference of $-3$ .

(Since

$9-12=-3$

$6-9=-3$

etc. Note that since the sequence is decreasing, the common difference is negative.)

To find the next $3$ terms, we just keep subtracting $3$ :

$\begin{array}{l}-6-3=-9\\ -9-3=-12\\ -12-3=-15\end{array}$

So, the next $3$ terms are $-9$ , $-12$ , and $-15$ .

To find a formula for the ${n}^{\text{th}}$ term, we substitute $n=1,{a}_{1}=12$ , and $d=-3$ in

${a}_{n}=dn+c$

to find $c$ .

$\begin{array}{l}12=-3\left(1\right)+c\\ c=15\end{array}$

So, a formula for the ${n}^{\text{th}}$ term of this sequence is

${a}_{n}=-3n+15$ .

Example 3:

$\left\{2,3,5,8,12,17,23,...\right\}$

is not an arithmetic sequence. The difference ${a}_{2}-{a}_{1}$ is $1$ , but the next difference ${a}_{3}-{a}_{2}$ is $2$ .

No formula of the form

${a}_{n}=dn+c$ can be written for this sequence.