## Algebra 2 (1st Edition)

$a_n=5(-\dfrac{14}{5})^{n-1}$ and $a_7= 2409.45$
The given series shows a geometric series with a common ratio $5r=-14$ Thus, $r=-\dfrac{14}{5}$ The nth term is given by $a_n= a_1 r^{n-1}$ This gives: $a_n=5(-\dfrac{14}{5})^{n-1}$ Plug in $n=7$ $a_7=5(-\dfrac{14}{5})^{7-1} \approx 2409.45$ Hence, $a_n=5(-\dfrac{14}{5})^{n-1}$ and $a_7= 2409.45$