## Algebra 2 (1st Edition)

$a_n=7(-0.6)^{n-1}$ and $a_7=0.326592$
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, $a_1=7; r=\dfrac{a_2}{a_1}=\dfrac{-4.2}{7}=-0.6$ From equation (1), we have $a_n=7(-0.6)^{n-1}$ Plug in $n=7$ $a_7=7(-0.6)^{7-1}=0.326592$ Hence, $a_n=7(-0.6)^{n-1}$ and $a_7=0.326592$