Answer
$a_n=2 (\dfrac{3}{4})^{n-1}$ and $a_7=\dfrac{729}{2048}$
Work Step by Step
The nth term is given by $a_n= a_1 r^{n-1}$
Here, $a_1=2; r=\dfrac{a_2}{a_1}=\dfrac{3/2}{2}=\dfrac{3}{4}$
$a_n=(2) \times (\dfrac{3}{4})^{n-1}$
This gives: $a_n=2 (\dfrac{3}{4})^{n-1}$
Plug in $n=7$.
$a_7=2 (\dfrac{3}{4})^{7-1}=\dfrac{729}{2048}$
Hence, $a_n=2 (\dfrac{3}{4})^{n-1}$ and $a_7=\dfrac{729}{2048}$
