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)

$\frac{7}{81} (179 +10^{-20})$

Answer 2

Let S = 0.7 + 0.77 + 0.777 +...
$= \frac{7}{10}+\frac{77}{10^2}+\frac{777}{10^3} +... upto\ 20\ terms$
$= 7 \bigg[\frac{1}{10}+\frac{11}{10^2}+\frac{111}{10^3} +... upto\ 20\ terms \bigg]$
$=\frac{7}{9}\bigg[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000} +... upto\ 20\ terms \bigg]$
$=\frac{7}{9}\Bigg[\bigg(1-\frac{1}{10}\bigg)+\bigg(1-\frac{1}{10^2}\bigg)+\bigg(1-\frac{1}{10^3}\bigg)$
\hspace70mm +...+ upto\ 20\ terms]
$=\frac{7}{9}[(1+1+...+\ upto\ 20\ terms)$
\hspace30mm -\bigg(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3}+...+ upto\ 20\ terms \bigg) \Bigg]
=\frac{7}{9}\Bigg[20-\frac{\frac{1}{10}\bigg\\{1-\big(\frac{1}{10}\big)^{20}\bigg\\}}{1-\frac{1}{10}}\Bigg]
\hspace30mm \Bigg[ \begin{array} \ \because \displaystyle\sum_{ i = 1}^{ 20} = 20\ and\ sum\ of\ n\ terms\ of \\\ GP, S_n = \frac{a (1 - r^n)}{1 - r } when \ ( r < 1 ) \\\ \end{array} \Bigg ]
=\frac{7}{9}\Bigg[20-\frac{1}{9}\bigg\\{1-\bigg(\frac{1}{10}\bigg)^{20}\bigg\\}{}\Bigg]
$=\frac{7}{9}\Bigg[\frac{179}{9}+\frac{1}{9}\bigg(\frac{1}{10}\bigg)^{20}\Bigg] = \frac{7}{81}[179+(10)^{-20}]$