Question

The sum of $100$ terms of the series $0.9+0.09+0.009+..........$ will be equal to

• A. $1-{\left(\frac{1}{10}\right)}^{100}$
• B. $1+{\left(\frac{1}{10}\right)}^{106}$
• C. $1-{\left(\frac{1}{10}\right)}^{106}$
• D. $1+{\left(\frac{1}{10}\right)}^{100}$

Answer 1

The correct option is A

$1-{\left(\frac{1}{10}\right)}^{100}$

Explanation for the correct option:

Given, $0.9+0.09+0.009+...$

$$\begin{gathered} \Rightarrow 9\left[0.1+0.01+0.001+...\right] \\ \Rightarrow 9\left[\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...\right] \end{gathered}$$

$\frac{1}{10},\frac{1}{100},\frac{1}{1000},.......$ is a geometric progression with the first term $a=\frac{1}{10}$ and $r=\frac{1}{10}$.

Sum of geometric progression$=a\left[\frac{1-r^n}{1-r}\right]$ for, $r<1$

Thus, sum of $\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+.......$ $=\frac{1}{10}\left[\frac{1-{\left({\displaystyle \frac{1}{10}}\right)}^{100}}{{\displaystyle 1-\frac{1}{10}}}\right]$

$$\begin{gathered} =\frac{1}{10}\times \left[\frac{1-{\left({\displaystyle \frac{1}{10}}\right)}^{100}}{{\displaystyle \frac{9}{10}}}\right] \\ =\frac{1}{10}\times \frac{10}{9}\left[1-{\left(\frac{1}{10}\right)}^{100}\right] \\ =\frac{1}{9}\left[1-{\left(\frac{1}{10}\right)}^{100}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow 9\left[\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+........\right]=9\times \frac{1}{9}\left[1-{\left(\frac{1}{10}\right)}^{100}\right] \\ \Rightarrow 9\left[\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+........\right]=\left[1-{\left(\frac{1}{10}\right)}^{100}\right] \end{gathered}$$

Thus, $0.9+0.09+0.009+..........$ is equal to $\left[1-{\left(\frac{1}{10}\right)}^{100}\right]$.

Hence, option (A) is correct.