# Finding the ${n}^{\text{th}}$ Term of an Arithmetic Sequence

Given an arithmetic sequence with the first term ${a}_{1}$ and the common difference $d$ , the ${n}^{\text{th}}$ (or general) term is given by ${a}_{n}={a}_{1}+\left(n-1\right)d$ .

Example 1:

Find the ${27}^{\text{th}}$ term of the arithmetic sequence  $5,8,11,54,...$ .

${a}_{1}=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}d=8-5=3$

So,

$\begin{array}{l}{a}_{27}=5+\left(27-1\right)\left(3\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=83\end{array}$

Example 2:

Find the ${40}^{\text{th}}$ term for the arithmetic sequence in which
${a}_{8}=60$ and ${a}_{12}=48$ .

Substitute $60$ for ${a}_{8}$ and $48$ for ${a}_{12}$ in the formula
${a}_{n}={a}_{1}+\left(n-1\right)d$  to obtain a system of linear equations in terms of ${a}_{1}$ and $d$ .

$\begin{array}{l}{a}_{8}={a}_{1}+\left(8-1\right)d\to 60={a}_{1}+7d\\ {a}_{12}={a}_{1}+\left(12-1\right)d\to 48={a}_{1}+11d\end{array}$

Subtract the second equation from the first equation and solve for $d$ .

$\begin{array}{l}12=-4d\\ -3=d\end{array}$

Then $60={a}_{1}+7\left(-3\right)$ .  Solve for $a$ .
$\begin{array}{l}60={a}_{1}-21\\ 81={a}_{1}\end{array}$

Now use the formula to find ${a}_{40}$ .

${a}_{40}=81+39\left(-3\right)=81-117=-36$ .