## College Algebra (11th Edition)

$a_8=-31 \text{ and } a_n=1-4n$
$\bf{\text{Solution Outline:}}$ To find $a_8$ and $a_n$ in the given sequence \begin{array}{l}\require{cancel} -3,-7,-11,... ,\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=-7-(-3) \\\\ d=-7+3 \\\\ d=-4 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_1=-3$ and $d=-4$ then \begin{array}{l}\require{cancel} a_n=-3+(n-1)(-4) \\\\ a_n=-3-4n+4 \\\\ a_n=1-4n .\end{array} With $n=8,$ then \begin{array}{l}\require{cancel} a_8=1-4(8) \\\\ a_8=1-32 \\\\ a_8=-31 .\end{array} Hence, $a_8=-31 \text{ and } a_n=1-4n .$