8. Sequences and Series
medium

શ્રેણી $0.7,0.77,0.777, . . . $ પ્રથમ $20$ પદોનો સરવાળો મેળવો.

A

$\frac{7}{{18}}\left( {179 - {{10}^{ - 20}}} \right)$

B

$\;\frac{7}{9}\left( {99 - {{10}^{ - 20}}} \right)$

C

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

D

$\;\frac{7}{9}\left( {99 + {{10}^{ - 20}}} \right)$

(JEE MAIN-2013)

Solution

$\frac{7}{10}+\frac{77}{100}+\frac{777}{10^{3}}+\ldots \ldots .+u p$ to $20$ terms

$=7\left[\frac{1}{10}+\frac{11}{100}+\frac{111}{10^{3}}+\ldots . . up \text { to } 20 \text { terms }\right]$

$=\frac{7}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots . .up \text { to } 20 \text { terms }\right]$

$=\frac{7}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\ldots . \text { up to } 20 \text { terms }\right]$

$=\frac{7}{9}\left[20-\frac{\frac{1}{10}\left(1-\left(\frac{1}{10}\right)^{20}\right)}{1-\frac{1}{10}}\right]=\frac{7}{9}\left[20-\frac{1}{9}\left(1-\left(\frac{1}{10}\right)^{20}\right)\right]$

$=\frac{7}{9}\left[\frac{179}{9}+\frac{1}{9}\left(\frac{1}{10}\right)^{20}\right]=\frac{7}{81}\left[179+(10)^{-20}\right]$

Standard 11
Mathematics

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