## Precalculus (6th Edition) Blitzer

$a_n=2n+1$
We know that $a_n=a_1+(n-1) d$ Here, we have $a_{3}=a_1+2d=7; a_{8}=a_1+7d=17$ Thus,we have $a_1+7d-a_1-2d=17-7$ or, $5d=10 \implies d=2$ and $a_1=7-2(2)=3$ Now, $a_n=3+(n-1) (2)=2n+1$