Question

How many terms of the 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

$\frac{2}{9},\frac{-1}{3},\frac{1}{2},..........$

$a=\frac{2}{9};r=\frac{\frac{-1}{3}}{\frac{2}{9}}=\frac{-3}{2}$

Sum of terms $=\frac{a(r^n-1)}{r-1}\frac{55}{72}=\frac{\frac{2}{9}({\left(\frac{-3}{2}\right)}^n-1)}{\frac{-3}{2}-1}\frac{55}{72}\times \frac{9}{2}=\frac{{\left(\frac{-3}{2}\right)}^n-1}{\frac{-5}{2}}\frac{55}{16}\times \frac{-5}{2}={\left(\frac{-3}{2}\right)}^n-1\frac{-275}{32}+1={\left(\frac{-3}{2}\right)}^n{\left(\frac{-3}{2}\right)}^n=\frac{-243}{32}{\left(\frac{-3}{2}\right)}^n={\left(\frac{-3}{2}\right)}^{5}n=5$