Answer
$a_{1} = -3, r= -4$
Work Step by Step
Given,
$a_{3} = -48$
$a_{4} = 192$
In geometric sequence, common ratio,
$r = \frac{a_{n}}{a_{n-1}}$
$r = \frac{a_{4}}{a_{3}} = \frac{192}{-48} =-4$
To find $a_{1},$
$a_{3} = a_{1} . r^{3-1} $ using $a_{n} = a_{1} . r^{n-1} $
$a_{3} = a_{1} . r^{2} $
Using $r$ and $a_{3}$ values,
$-48= a_{1} .(-4)^{2} $
$-48 = a_{1} .(16) $
$a_{1} = \frac{-48}{16}$
$a_{1} = -3$
