Answer
See below.
Work Step by Step
$a_1=0$
$a_2=a_1+3=0+3=3$
$a_3=a_2+3=3+3=6$
$a_4=a_3+3=6+3=9$
$a_5=a_4+3=9+3=12$
$a_6=a_5+3=12+3=15$
The first differences: $a_2-a_1=a_3-a_2=a_4-a_3=a_5-a_4=a_6-a_5=3$
The second differences: $3-3=3-3=3-3=3-3=0$.
The first differences are constant; thus it is linear.
