## Algebra 2 (1st Edition)

$a_{1}=4$ $a_{2}=28$ $a_{3}=196$ $a_{4}=1372$ $a_{5}=9604$ $a_{6}=67,228$ $a_{7}=470,596$ $a_{8}=3,294,172$ The sequence is geometric.
$a_{1}=4$ $a_{2}=7a_{1}=28$ $a_{3}=7a_{2}=196$ $a_{4}=7a_{3}=1372$ $a_{5}=7a_{4}=9604$ $a_{6}=7a_{5}=67,228$ $a_{7}=7a_{6}=470,596$ $a_{8}=7a_{7}=3,294,172$ The consecutive terms are such that $\displaystyle \frac{a_{n}}{a_{n-1}}=\frac{7a_{n-1}}{a_{n-1}}=7$ A common ratio exists, so the sequence is geometric.