## Precalculus (10th Edition)

The common difference is: $d=6.$ The initial term: $a_1=-53.$ $a_n=-59+6n$, $a_n=a_{n-1}+6$
We know that $a_{9}=-5,a_{15}=31$. Thus the common difference is: $d=\frac{a_k-a_l}{k-l}=\frac{a_{15}-a_{}}{15-9}=\frac{31-(-5)}{6}=6.$ The initial term: $a_1=a_n-(n-1)d=a_9-(8)d=-5-(8)6=-53.$ Thus: $a_n=a_1+(n-1)d=-53+(n-1)6=-59+6n$, $a_n=a_{n-1}+d=a_{n-1}+6$