EPFO 2017 English Medium Solved Paper

Given, initial quantity of hydrogen gas in container = 300 liter
First day loss in quantity = 100 liter
Second day loss in quantity = $\frac{100}{3}\mathrm{liter}$$=\frac{100}{3^1}$
Third day loss in quantity= $\frac{100}{9}\mathrm{liter}=\frac{100}{3^2}$
Fourth day loss in quantity$=\frac{100}{27}\mathrm{liter}=\frac{100}{3^3}$
∴ 100th day loss in quantity $=\frac{100}{3^{99}}\mathrm{liter}.$
Total loss of gas
$=100+\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}$ $+....+\frac{100}{3^{99}}$
$=100\left[1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....\frac{1}{3^{99}}\right]$
Series $1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{99}}$ is a
geometric progression. Here,a= 1 and r =$\frac{1}{3}$
Sum of geometric progression = $\frac{\mathit{a}\left(1-{\mathit{r}}^{\mathit{n}}\right)}{\left(1-\mathit{r}\right)}$
${\mathit{S}}_{100}=100\left[1\frac{\left(1-\frac{1}{3^{100}}\right)}{1-\frac{1}{3}}\right]$ $=100\left[\frac{3^{100}-1}{\frac{2}{3}\times 3^{100}}\right]$
$\frac{100\times 3}{2}$ $\left[\frac{3^{100}-1}{3^{100}}\right]$
=$150\times \left(\frac{3^{100}-1}{3^{100}}\right)$
$\because \frac{3^{100}-1}{3^{100}}$<1
Hence, the container process at least 150 liter of hydrogen gas on 100th day.