The sum of first 20 terms of the sequence&nbs | vedclass

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Question

$\frac{7}{10}+\frac{77}{100}+\frac{777}{10^{3}}+\ldots \ldots .+u p$ to $20$ terms

$=7\left[\frac{1}{10}+\frac{11}{100}+\frac{111}{10^{3}}+\ldots .

Vedclass · Accepted Answer

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

Vedclass · Answer

$\frac{7}{10}+\frac{77}{100}+\frac{777}{10^{3}}+\ldots \ldots .+u p$ to $20$ terms

$=7\left[\frac{1}{10}+\frac{11}{100}+\frac{111}{10^{3}}+\ldots . . up 	ext { to } 20 	ext { terms }ight]$

$=\frac{7}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots . .up 	ext { to } 20 	ext { terms }ight]$

$=\frac{7}{9}\left[\left(1-\frac{1}{10}ight)+\left(1-\frac{1}{10^{2}}ight)+\left(1-\frac{1}{10^{3}}ight)+\ldots . 	ext { up to } 20 	ext { terms }ight]$

$=\frac{7}{9}\left[20-\frac{\frac{1}{10}\left(1-\left(\frac{1}{10}ight)^{20}ight)}{1-\frac{1}{10}}ight]=\frac{7}{9}\left[20-\frac{1}{9}\left(1-\left(\frac{1}{10}ight)^{20}ight)ight]$

$=\frac{7}{9}\left[\frac{179}{9}+\frac{1}{9}\left(\frac{1}{10}ight)^{20}ight]=\frac{7}{81}\left[179+(10)^{-20}ight]$

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

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