$\begin{array}{1 1}2059 \\ 665 \\ \large\frac{2059}{3} \\ 1995 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- $n^{th}$ term of a G.P.= $t_n=a.r^{n-1}$

Given that in a G.P.first term $a=729$,and $7^{th}$ term $=t_7=64$

We know that $t_n=a.r^{n-1}$

$\therefore t_7=64=a.r^{7-1}$

$\Rightarrow\:64=729\times r^6$

But $64=2^6$ and $729=3^6$

$\Rightarrow\:r^6=\large\frac{64}{729}=\frac{2^6}{3^6}=\big(\frac{2}{3}\big)^6$

$\therefore\:r=\large\frac{2}{3}$

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

$\therefore S_7=729.\large\frac{1-(2/3)^7}{1-2/3}$

$=729\times 3\times \big(1-(\large\frac{2}{3})^7\big)$

$=3^7\times\large\frac{ (3^7-2^7)}{3^7}$

$=2187-128=2059

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...