## Intermediate Algebra (6th Edition)

$a_{n}=4n-1$
Given Sequence: $3,7,11,15$ We can write it as $4-1,8-1,12-1,16-1$ $a_{1} = 3 = 4-1 = 4(1)-1$ $a_{2} = 7 = 8-1 = 4(2)-1$ $a_{3} = 11 = 12-1 = 4(3)-1$ $a_{4} = 15 = 16-1 = 4(4)-1$ General term : $a_{n}=4(n)-1$