Question

The sum of first 20 terms of the sequence 0.3, 0.33, 0.333, ....., is

• A. $\frac{1}{27}[189+(10{)}^{-20}]$
• B. $\frac{1}{27}[179+(10{)}^{-20}]$
• C. $\frac{1}{27}[149+(10{)}^{-20}]$
• D. $\frac{1}{27}[255+(10{)}^{-20}]$

Answer 1

The correct option is B

$\frac{1}{27}[179+(10{)}^{-20}]$

Let $S=0.3+0.33+0.333+\dots \dots$
$=\frac{3}{10}+\frac{33}{{10}^2}+\frac{333}{{10}^3}+\dots \dots \text{upto 20 terms}$
$=3[\frac{1}{10}+\frac{11}{{10}^2}+\frac{111}{{10}^3}+\dots \dots \text{upto 20 terms}]$
$=\frac{3}{9}[(1-\frac{1}{10})+(1-\frac{1}{{10}^2})+(1-\frac{1}{{10}^3})+\dots \dots +\text{upto 20 terms}]$
$=\frac{3}{9}\left[\right(1+1+\dots \dots +\text{upto 20 terms})-(\frac{1}{10}+\frac{1}{{10}^2}+\frac{1}{{10}^3}+\dots \dots +\text{upto 20 terms})]$
$=\frac{3}{9}[20-\frac{\frac{1}{10}\{1-{\left(\frac{1}{10}\right)}^{20}\}}{1-\frac{1}{10}}]$
$(\because S_n=\frac{a(1-r^n)}{1-r}\text{when}r<1)$
$=\frac{3}{9}[20-\frac{1}{9}\{1-{\left(\frac{1}{10}\right)}^{20}\}]$
$=\frac{3}{9}[\frac{179}{9}+\frac{1}{9}{\left(\frac{1}{10}\right)}^{20}]=\frac{3}{81}[179+(10{)}^{-20}]$

$=\frac{1}{27}[179+(10{)}^{-20}]$