Answer
$a_{1} = \frac{4}{9}, r= -3$
Work Step by Step
Given,
$a_{3} = 4$
$a_{4} = -12$
In geometric sequence, common ratio,
$r = \frac{a_{n}}{a_{n-1}}$
$r = \frac{a_{4}}{a_{3}} = \frac{-12}{4} =-3$
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,
$4= a_{1} .(-3)^{2} $
$4 = a_{1} .(9) $
$a_{1} = \frac{4}{9}$
