Answer
$a_{1} = \frac{2}{3}, r= -2$
Work Step by Step
Given,
$a_{2} = -\frac{4}{3}$
$a_{3} = \frac{8}{3}$
In geometric progression, common ratio,
$r = \frac{a_{n}}{a_{n-1}}$
$r = \frac{a_{3}}{a_{2}} = \frac{\frac{8}{3}}{-\frac{4}{3}} $
$= \frac{8}{3} \times -\frac{3}{4} = -2$
To find $a_{1},$
$a_{2} = a_{1} . r^{2-1} $ using $a_{n} = a_{1} . r^{n-1} $
$a_{2} = a_{1} . r^{1} $
Using $r$ and $a_{2}$ values,
$-\frac{4}{3}= a_{1} .(-2)^{1} $
$-\frac{4}{3} = a_{1} .(-2) $
$a_{1} = -\frac{4}{3} \times \frac{1}{-2}$
$a_{1} =\frac{2}{3}$
