Question

The sum of first 20 terms of the sequence 0.5, 0.55, 0.555,...., is ___.

• A. $\frac{5}{81}(179-{10}^{-20})$
• B. $\frac{5}{9}(99-{10}^{-20})$
• C. $\frac{5}{81}(179+{10}^{-20})$
• D. $\frac{5}{9}(99+{10}^{-20})$

Answer 1

The correct option is C $\frac{5}{81}(179+{10}^{-20})$

$\text{Let S}=0.5+0.55+0.555+...\text{upto 20 terms}=\frac{5}{10}+\frac{55}{100}+\frac{555}{{10}^3}+...+\text{upto 20 terms}$
$=5[\frac{1}{10}+\frac{11}{100}+\frac{111}{{10}^3}+...+\text{upto 20 terms}]$
Multiplying and dividing the RHS by 9, we get
$S=\frac{5}{9}[\left(\right(1-\frac{1}{10})+(1-\left(\frac{1}{10}{)}^2\right)+(1-(\frac{1}{10}{)}^3\left)\right)+.....\text{upto 20 terms}]=\frac{5}{9}[20-\frac{\frac{1}{10}(1-(\frac{1}{10}{)}^{20})}{1-\frac{1}{10}}]=\frac{5}{9}[\frac{179}{9}+\frac{1}{9}(\frac{1}{10}{)}^{20}]=\frac{5}{81}[179+(10{)}^{-20}]$