Question

The sum of $100$ terms of the series $.9+09+.009$. will be?

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

$0.9+0.09+0.009.......$

$=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+.......$

$=\frac{9}{10}+\frac{9}{{10}^2}+\frac{9}{{10}^3}.........$

The series is in G.P. , where

first term, $a=\frac{9}{10}$

common ratio, $r=\frac{1}{10}$

Sum of n terms, $S_n=\frac{a(1-r^n)}{1-r}$

So, sum of $100$ terms,

$S_{100}=\frac{\frac{9}{10}(1-(\frac{1}{10}{)}^{100})}{1-\frac{1}{10}}$

$S_{100}=\frac{\frac{9}{10}(1-(\frac{1}{10}{)}^{100})}{\frac{9}{10}}$

$S_{100}=1-(\frac{1}{10}{)}^{100}$