Question

The sum of 10 terms of the series $0.7+.77+.777+\dots \dots \dots$ is-

• A. $\frac{7}{9}\left(89+\frac{1}{10^{10}}\right)$
• B. $\frac{7}{81}\left(89+\frac{1}{10^{10}}\right)$
• C. $\frac{7}{81}\left(89+\frac{1}{10^9}\right)$
• D. $\frac{7}{9}\left(89+\frac{1}{10^9}\right)$

Answer 1

Answer: (B)

The sum of 10 terms of series $.7+.77+.777+\dots \dots$
$S_{10}=+\frac{7}{10}+\frac{77}{100}\dots \dots \dots$. to 10 terms
$=\frac{7}{9}[\frac{9}{10}+\frac{99}{100}+\dots \dots \dots$.to 10 terms $]$
$=\frac{7}{9}[\frac{10-1}{10}+\frac{100-1}{100}+\frac{1000-1}{1000}+\dots \dots \dots$ to 10 terms $]$
$=\frac{7}{9}[(1+1+1\dots \dots ..10$ terms $)-(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\dots \dots \dots .$. to 10 terms $)]$
$=\frac{7}{9}\left[10-\frac{\frac{1}{10}\left(1-{\left(\frac{1}{10}\right)}^{10}\right)}{1-\frac{1}{10}}\right]=\frac{7}{9}\left[10-\frac{\left(10^{10}-1\right)}{9.10^{10}}\right]=\frac{7}{81}\left[90-1+\frac{1}{10^{10}}\right]=\frac{7}{81}\left[89+\frac{1}{10^{10}}\right]$