Answer
$d=3$.
$a_1 =-13$,
$a_n= a_{n-1}+3$.
$a_n= 3n-16$
Work Step by Step
1. Based on the given conditions, we have $a_8=a_1+7d=8$ and $a_{20}=a_1+19d=44$, thus $12d=36$ and $d=3$.
2. We can find the first term $a_1=8-7(3)=-13$, and a recursive formula $a_n=a_{n-1}+d=a_{n-1}+3$.
3. We can find a formula for the nth term $a_n=a_1+(n-1)d=-13+(n-1)3=3n-16$
