Answer
$d=4$.
$a_1= 18$,
$a_n= a_{n-1}+4$.
$a_n= 4n+14$
Work Step by Step
1. Based on the given conditions, we have $a_{5}=a_1+4d=-2$ and $a_{13}=a_1+12d=30$, thus $8d=32$ and $d=4$.
2. We can find the first term $a_1=-2-4(4)=18$, and a recursive formula $a_n=a_{n-1}+d=a_{n-1}+4$.
3. We can find a formula for the nth term $a_n=a_1+(n-1)d=18+(n-1)(4)=4n+14$