Answer
$a_{2}$ = -6
$a_{3}$ = 18
Work Step by Step
We know nth term $a_{n}$ = $a_{1}r^{n-1}$
Given
Geometric sequence = 2, $a_{2}$, $a_{3}$, -54
$a_{1}$ = 2
$a_{4}$ = -54
From above
$a_{4}$ = $2r^{4-1}$ = $2 r^{3}$ = -54
$r^{3}$ = -27
r = $\sqrt[3] (-27)$ = -3
$a_{2}$ = $a_{1}r^{2-1}$
= $2(-3)^{2-1}$
= 2$\times$(-3)
= -6
$a_{3}$ = $a_{1} r^{3-1}$
= $2(-3)^{3-1}$
= $2 (-3)^{2}$
= 2$\times$9
= 18
