$\begin{array}{1 1} \large\frac{1-(0.1)^{20}}{6} \\\large\frac{1-(0.1)^{20}}{60} \\ \large\frac{1-(0.1)^{20}}{3} \\ \large\frac{1-(0.1)^{20}}{30} \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Sum of $n$ terms of a G.P$=S_n=a.\large\frac{1-r^n}{1-r}$

Given G.P. is $ 0.15+0.015+0.0015..........upto\:20\:terms$

In this G.P. $a=0.15\:\;and$ common ratio $r=\large\frac{0.015}{0.15}=\frac{1}{10}$$=0.1$

We know that sum of $n$ terms of a G.P.=$S_n=a.\large\frac{1-r^n}{1-r}$

$\therefore\:S_{20}=0.15\times \large\frac{1-(0.1)^{20}}{1-0.1}$

$=0.15\times\large\frac{1-(0.1)^{20}}{0.9}$

$=\large\frac{1}{6}$$\times (1-(0.1)^{20})$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...