Question

Find the sum of the first $25$ terms of the geometric series
$16-48+144-432+.....$

Answer 1

We are given that the sequence is a G.P

The firs term, $a=16$ and the common ratio, $r=-\frac{48}{16}=-3$

Sum of the $25$ terms $=a\left(\frac{1-r^{25}}{1-r}\right)$

$=16\left(\frac{1-{(-3)}^{25}}{1-(-3)}\right)$

$=4(1+3^{25})$