Question

The sum of first $20$ terms of the sequence $0.7, 0.77, 0.777,....,$ is

• A. $\frac{7}{81} (179 -10^{-20})$
• B. $\frac{7}{9} (99 -10^{-20})$
• C. $\frac{7}{81} (179 +10^{-20})$
• D. $\frac{7}{9} (99 +10^{-20})$

Answer 1

Answer: (C)

Let $S=0.7+0.77+0.777+\ldots$
$=\frac{7}{10}+\frac{77}{10^{2}}+\frac{777}{10^{3}}+\ldots$ upto $20$ terms
$=7\left[\frac{1}{10}+\frac{11}{10^{2}}+\frac{111}{10^{3}}+\ldots\right.$ upto $20$ terms $]$
$=\frac{7}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots\right.$ upto $20$ terms $]$
$=\frac{7}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)\right.$
$+\ldots+$ upto $20$ terms ]
$=\frac{7}{9}[(1+1+\ldots+$ upto $20$ terms $)$
$-\left(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\ldots+\right.$ upto $20$ terms $\left.)\right]$
$=\frac{7}{9}\left[20-\frac{\frac{1}{10}\left\{1-\left(\frac{1}{10}\right)^{20}\right\}}{1-\frac{1}{10}}\right]$
$\left[\because \displaystyle\sum_{i=1}^{20}=20\right.$ and sum of $n$ terms of $] G P, S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$ when $(r < 1)$
$ =\frac{7}{9}\left[20-\frac{1}{9}\left\{1-\left(\frac{1}{10}\right)^{20}\right\}\right] $
$=\frac{7}{9}\left[\frac{179}{9}+\frac{1}{9}\left(\frac{1}{10}\right)^{20}\right]=\frac{7}{81}\left[179+(10)^{-20}\right] $