Sequences and Series 4 Question 6
6.
The sum of first 20 terms of the sequence $0.7,0.77,0.777, \ldots$, is
(a) $\frac{7}{81}\left(179-10^{-20}\right)$
(b) $\frac{7}{9}\left(99-10^{-20}\right)$
(c) $\frac{7}{81}\left(179+10^{-20}\right)$
(d) $\frac{7}{9}\left(99+10^{-20}\right)$
(2013 Main)
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Answer:
Correct Answer: 6. (c)
Solution:
- Let $S=0.7+0.77+0.777+\ldots$
$=\frac{7}{10}+\frac{77}{10^{2}}+\frac{777}{10^{3}}+\ldots$ upto 20 terms
$=7 [\frac{1}{10}+\frac{11}{10^{2}}+\frac{111}{10^{3}}+\ldots$ upto 20 terms]
$=\frac{7}{9} [\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots$ upto 20 terms]
$=\frac{7}{9} \quad [(1-\frac{1}{10})+(1-\frac{1}{10^{2}})+(1-\frac{1}{10^{3}})$ +… +upto 20 terms]
$ =\frac{7}{9}[(1+1+\ldots+\text { upto } 20 \text { terms }) $
$ -\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\ldots+\text { upto } 20 \text { terms } ]$
$ =\frac{7}{9}[ 20-\frac{\frac{1}{10} 1-\frac{1}{10}}{1-\frac{1}{10}}] $
$ [ \sum _{i=1}^{20}=20 \text { and sum of } n \text { terms of GP ,} S_n=\frac{a\left(1-r^{n}\right)}{1-r} \text { when }(r<1) ] $
$ =\frac{7}{9} [\frac{179}{9}+\frac{1}{9} (\frac{1}{10})^{20} ]$
$=\frac{7}{81}[179+(10)^{-20}]$
