Answer
$a_n=4 (\dfrac{1}{2})^{n-1}$ and $a_7=\dfrac{1}{16}$
Work Step by Step
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1)
Here, $a_1=4; r=\dfrac{a_2}{a_1}=\dfrac{2}{4}=\dfrac{1}{2}$
From equation (1), we have
$a_n=4 (\dfrac{1}{2})^{n-1}$
Plug in $n=7$
$a_7=4 (\dfrac{1}{2})^{7-1}=\dfrac{1}{16}$
Hence, $a_n=4 (\dfrac{1}{2})^{n-1}$ and $a_7=\dfrac{1}{16}$
