## Precalculus (6th Edition) Blitzer

$a_1=-2;a_2 =3, a_3=8; a_4=13; a_5=18: a_6=23$
We need to use the formula $a_n =a_1+(n-1) d$ ....(1) Here, $a_1=-2$ and $n=6$ and $d=1$ $a_2=-2 +(6-1) \times 1 =-2+5=3$ We make use of the recursive formula to get higher order terms. So, the first $6$ terms are: $a_1=-2;a_2 =3, a_3=8; a_4=13; a_5=18: a_6=23$