## College Algebra (11th Edition)

$a_8=x+21 \text{ and } a_n=x+3n-3$
$\bf{\text{Solution Outline:}}$ To find $a_8$ and $a_n$ in the given sequence \begin{array}{l}\require{cancel} a_{1}=x, a_{2}=x+3 ,\end{array} use the formula for finding the $n$th term of an arithmetic sequence. $\bf{\text{Solution Details:}}$ Since the common difference, $d,$ is the difference between a term and the term preceeding it, then \begin{array}{l}\require{cancel} d=a_{2}-a_{1} \\\\ d=x+3-x \\\\ d=3 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_{1}=x,$ $d=3,$ then \begin{array}{l}\require{cancel} a_n=x+(n-1)3 \\\\ a_n=x+3n-3 .\end{array} With $n=8,$ then \begin{array}{l}\require{cancel} a_8=x+3(8)-3 \\\\ a_8=x+24-3 \\\\ a_8=x+21 .\end{array} Hence, $a_8=x+21 \text{ and } a_n=x+3n-3 .$