Question

Find how many terms of G.P.
$\frac{2}{9}-\frac{1}{3}+\frac{1}{2}...$ must be added to get the sum equal to $\frac{55}{72}$ ?

Answer 1

$a_1=\frac{2}{9};a_2=\frac{-1}{3},a_3=\frac{1}{2}$

$\therefore r=\frac{a_2}{a_1}=\frac{-\frac{1}{3}}{\frac{2}{9}}=\frac{-1}{\left(3\right)\left(2\right)}\times 9$

$r=-\frac{3}{2}$

We know that sum of n terms of a G.P. = $\frac{a(1-r^n)}{(1-r)}$

$\therefore$ Sum = $\frac{2}{9}\frac{(1-(\frac{-3}{2}{)}^n)}{1-(-\frac{3}{2})}$

$\therefore \frac{55}{72}=\frac{2}{9}\times \frac{2}{5}\times (1-{\left(\frac{-3}{2}\right)}^n)$

$\therefore \frac{55\times 5}{8\times 4}=(1-{\left(\frac{-3}{2}\right)}^n)$

$\therefore \frac{275}{32}=1-{\left(\frac{-3}{2}\right)}^n$

$\therefore {\left(\frac{-3}{2}\right)}^n=\frac{1-275}{32}$

$\therefore {\left(\frac{-3}{2}\right)}^n=-\frac{243}{32}$

$\therefore n=5$

$\therefore$ Number of term is 5