Question

Sum of a certain number of terms of the series $\frac{2}{9},-\frac{1}{3},\frac{1}{2}....$ is $\frac{55}{72}$; Find the number.

Answer 1

Here $a=\frac{2}{9}$ and $r=-\frac{1}{3}.\frac{9}{2}=-\frac{3}{2}$
Let the number of terms be $n$.
Also $S_n=\frac{a(1-r^n)}{1-r}=\frac{55}{72}$
$\therefore \frac{2}{9},\frac{[1-(-3/2{)}^n]}{1-(-3/2)}=\frac{55}{72}$
or $1-{(-\frac{3}{2})}^n=\frac{55}{72}\times \frac{9}{2}\times \frac{5}{2}=\frac{275}{32}$
$1-\frac{275}{32}={(-\frac{3}{2})}^n$ or $-\frac{243}{32}={(-\frac{3}{2})}^n$
or ${(-\frac{3}{2})}^5={\left(\frac{3}{2}\right)}^n$ $\therefore n=5$