Answer
$1$
Work Step by Step
Given sequence,
$\frac{3}{2},-\frac{3}{4},\frac{3}{8}...$
$a_{1} = \frac{3}{2}$
$r = \frac{a_{2}}{a_{1}}$
$r = \frac{-\frac{3}{4}}{\frac{3}{2}}$
$r = -\frac{3}{4} \times \frac{2}{3}$
$r = -\frac{1}{2}$
Sum of the terms of an infinite geometric sequence
$S_{∞} = \frac{a_{1}}{1-r}$
Substituting $a_{1}$and $r$ values
$S_{∞} = \frac{\frac{3}{2}}{1-(-\frac{1}{2})}$
$S_{∞} = \frac{\frac{3}{2}}{1+\frac{1}{2}}$
$S_{∞} = \frac{\frac{3}{2}}{\frac{3}{2}}$
$S_{∞} = 1$
