## College Algebra (11th Edition)

$a_1=3$ $a_2=9$ $a_3=15$ $a_4=21$ $a_5=27$
Sequence parameter. $a_n=6n-3$ In order to find the nth term of a sequence, substitute n into the sequence parameter. Thus, to find the first 5 terms, we substitute n =1, 2, 3, 4, 5. $a_1=6(1)-3=3$ $a_2=6(2)-3=9$ $a_3=6(3)-3=15$ $a_4=6(4)-3=21$ $a_5=6(5)-3=27$