# 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:
**

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

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:
**

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

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:
**

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

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.

See also geometric sequences .