## Algebra 2 (1st Edition)

$a_n=6(3)^{n-1}$ and $a_7=4374$
Here, we have $6,18,54, 162$ This shows a geometric series with a common ratio of $6r=18$ Thus, $r=3$ The nth term is given by $a_n= a_1 r^{n-1}$ This gives: $a_n=(6) \times (3)^{n-1}$ Plug in $n=7$ $a_7=6 \times (3)^{7-1}=4374$ Hence, $a_n=6(3)^{n-1}$ and $a_7=4374$