Answer
$a_n=80 (-\dfrac{1}{2})^{n-1}$
Work Step by Step
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1)
Here, we have $a_2=a_1 r \implies r=\dfrac{-40}{a_1}$
and $a_4=a_1 r^{4-1} \implies r=\dfrac{-40}{a_1}$
Equation (1) gives: $-10=a_1 \times (\dfrac{-40}{a_1})^3$
$-10=a_1 \times \dfrac{-64000}{a_1^3}$
This gives: $a_1=\sqrt {6400}=80$
Hence, $a_n=80 (-\dfrac{1}{2})^{n-1}$
