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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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]
$$ \begin{aligned} & =\frac{7}{9}[(1+1+\ldots+\text { upto } 20 \text { terms }) \\ & -\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\ldots+\text { upto } 20 \text { terms } \\ & =\frac{7}{9} 20-\frac{\frac{1}{10} 1-\frac{1}{10}}{1-\frac{1}{10}} \\ & =\frac{7}{9} 20 \sum _{i=1}^{20}=20 \text { and sum of } n \text { terms of } \\ & =\frac{7}{9} \frac{179}{9}+\frac{1}{9} \frac{1}{10} \quad \frac{1}{20} \frac{1}{10}^{20}=\frac{a\left(1-r^{n}\right)}{1-r} \text { when }(r<1) \\ & \left.=179+(10)^{-20}\right] \end{aligned} $$
