Question

The sum of first $20$ terms of the sequence$0.7,0.77,0.777,...$, 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. $\frac{7}{9}\left(99+{10}^{-20}\right)$

Answer 1

The correct option is C

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

Explanation for correct option:

Let ,

$$\begin{gathered} S_n=0.7+0.77+0.777+.....+0.777777...(20\mathrm{times}) \\ \Rightarrow S_n=7\left(0.1+0.11+0.111+.....+0.111111(20\mathrm{times})\right) \\ \Rightarrow S_n=\frac{7}{9}\left(0.9+0.99+0.999+...+0.9999999(20\mathrm{times})\right) \\ \Rightarrow S_n=\frac{7}{9}\left[\left(1-0.1\right)+\left(1-0.01\right)+\left(1-0.001\right)+....+\left(1-0.000..01\right)\right] \\ \Rightarrow S_n=\frac{7}{9}\left[20-\left(\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+...+\frac{1}{{10}^{20}}\right)\right] \end{gathered}$$

We know that, the sum of the G.P$=\frac{a\left(1-r^n\right)}{1-r}$

$$\begin{gathered} \Rightarrow S_n=\frac{7}{9}\left[20-\frac{1}{10}\left(\frac{1-{\displaystyle \frac{1}{{10}^{20}}}}{1-{\displaystyle \frac{1}{10}}}\right)\right] \\ \Rightarrow S_n=\frac{7}{9}\left[20-\left(\frac{1-{\displaystyle \frac{1}{{10}^{20}}}}{9}\right)\right] \\ \Rightarrow S_n=\frac{7}{9}\left[20-\frac{1}{9}\left(1-{10}^{-20}\right)\right] \\ \Rightarrow S_n=\frac{7}{81}\left[179+{10}^{-20}\right] \end{gathered}$$

Hence, option C is correct.