Question

Find the sum of the sequence $96-48+24...$ upto $10$ terms.

Answer 1

The series is

$96-48+\mathrm{24.......}$

Here,

$\frac{a_2}{a_1}=\frac{-48}{96}=-\frac{1}{2}$

$\frac{a_3}{a_2}=\frac{24}{-48}=-\frac{1}{2}$

So the series is in G.P. with common ratio, $r=-\frac{1}{2}$

$S_{10}=\frac{a(1-r^{10})}{1-r}$

$S_{10}=\frac{96(1-{(-\frac{1}{2})}^{10})}{1+\frac{1}{2}}$

$=96\times \frac{2}{3}\times \left(\frac{2^{10}-1}{2^{10}}\right)$

$=64\left(\frac{2^{10}-1}{2^{10}}\right)$

$=\frac{2^{10}-1}{2^4}$

$=\frac{1023}{16}$