## College Algebra (11th Edition)

$a_8=-3 \text{ and } a_n=4.5n-39$
$\bf{\text{Solution Outline:}}$ To find $a_8$ and $a_n$ in the given sequence \begin{array}{l}\require{cancel} a_{10}=6, a_{11}=10.5 ,\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_{11}-a_{10} \\\\ d=10.5-6 \\\\ d=4.5 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_{10}=6$ $d=4.5,$ and $n=10,$ then \begin{array}{l}\require{cancel} a_{10}=a_1+(10-1)d \\\\ 6=a_1+(10-1)4.5 \\\\ 6=a_1+(9)4.5 \\\\ 6=a_1+40.5 \\\\ 6-40.5=a_1 \\\\ a_1=-34.5 .\end{array} Using $a_n=a_1+(n-1)d$ with $a_{1}=-34.5$ and $d=4.5,$ then \begin{array}{l}\require{cancel} a_n=-34.5+(n-1)4.5 \\\\ a_n=-34.5+4.5n-4.5 \\\\ a_n=4.5n-39 .\end{array} With $n=8,$ then \begin{array}{l}\require{cancel} a_8=4.5(8)-39 \\\\ a_8=36-39 \\\\ a_8=-3 .\end{array} Hence, $a_8=-3 \text{ and } a_n=4.5n-39 .$