Answer
$a_{2}$ = 12
$a_{3}$ = 18
Work Step by Step
We know nth term $a_{n}$ = $a_{1}r^{n-1}$
Given
Geometric sequence = 8, $a_{2}$, $a_{3}$, 27
$a_{1}$ = 8
$a_{4}$ = 27
From above
$a_{4}$ = $8 r^{4-1}$ = $8 r^{3}$ = 27
r = $\sqrt[3] \frac{27}{8}$ = $\frac{3}{2}$
$a_{2}$ = $a_{1}r^{2-1}$
= $8 (\frac{3}{2})$
= 12
$a_{3}$ = $a_{1} r^{3-1}$
= $8 (\frac{3}{2})^{2}$
= 8$\times$$\frac{9}{4}$
= 18
