## Algebra and Trigonometry 10th Edition

$a_1=0$ $a_2=a_1+3=0+3=3$ $a_3=a_2+3=3+3=6$ $a_4=a_3+3=6+3=9$ $a_5=a_4+3=9+3=12$ $a_6=a_5+3=12+3=15$ 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.