#### Answer

$a_n=3 (\dfrac{-2}{5})^{n-1}$ and $a_7=\dfrac{192}{15625}$

#### Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$
Here, $a_1=3; r=\dfrac{a_2}{a_1}=\dfrac{-6/5}{2}=\dfrac{-2}{5}$
$a_n=(3) \times (\dfrac{-2}{5})^{n-1}$
This gives: $a_n=3 (\dfrac{-2}{5})^{n-1}$
Plug in $n=7$
$a_7=3 (\dfrac{-2}{5})^{7-1}=\dfrac{192}{15625}$
Hence, $a_n=3 (\dfrac{-2}{5})^{n-1}$ and $a_7=\dfrac{192}{15625}$