Answer
$a_{4}= - \frac{8}{9}$
Work Step by Step
Given,
$a_{1} = 3$
$r = -\frac{2}{3}$
$a_{n} $ of geometric sequence is $a_{n} = a_{1} . r^{n-1} $
To find fourth term, substitute $a_{1} ,$ $r$ values and $n=4$
$a_{4} = a_{1} . r^{4-1} $
$a_{4} = a_{1} . r^{3} $
$a_{4} = 3 \times (-\frac{2}{3})^{3} $
$a_{4} = 3 \times (-\frac{8}{27})$
$a_{4} = -\frac{8}{9}$
