Answer
$a_n=-2(0.4)^{n-1}$ and $a_7=-0.008192$
Work Step by Step
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1)
Here, $a_1=-2; r=\dfrac{a_2}{a_1}=\dfrac{-0.8}{-2}=0.4$
From equation (1), we have
$a_n=-2(0.4)^{n-1}$
Plug in $n=7$
$a_7=-2(0.4)^{7-1}=-0.008192$
Hence, $a_n=-2(0.4)^{n-1}$ and $a_7=-0.008192$
