## Intermediate Algebra (6th Edition)

$a_{5} = 648$
$a_{1} = 8$ $r = -3$ $a_{n}$ of geometric sequence is $a_{n} = a_{1} . r^{n-1}$ To find fifth term, substitute $a_{1} ,$ $r$ values and $n=5$ $a_{5} = a_{1} . r^{5-1}$ $a_{5} = a_{1} . r^{4}$ $a_{5} = 8 \times(-3)^{4}$ $a_{5} = 8 \times 81$ $a_{5} = 648$