$\begin{array}{1 1}2 \\ 4 \\ 8 \\ 6 \end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Since it is given that the number of terms is even number,

let the number of terms in the G.P be $2n$

$\therefore$ The G.P. is $a,ar,ar^2,.........a.r^{2n-1}$......(i)

The terms occupying odd places form the sequence

$a,ar^2,ar^4,.......ar^{2n-2}$..........(ii)

which is also a G.P. with $1^{st}$ term $=a$ and common ration $=r^2$

and no. of terms $=n$

It is given that $(a+ar+ar^2+.......ar^{2n-1})=5(a+ar^2+ar^4+........a.r^{2n-2})$

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

$\Rightarrow\:a+ar+ar^2+.......a.r^{2n-1}=a.\large\frac{r^{2n}-1}{r-1}$ and

$a+ar^2+ar^4+........a.r^{2n-2}= a.\large\frac{(r^2)^{n}-1}{r^2-1}$$=a.\large\frac{r^{2n}-1}{(r-1)(r+1)}$

$\Rightarrow\: a.\large\frac{(r)^{2n}-1}{r-1}$$=5\times a.\large\frac{(r)^{2n}-1}{(r-1)(r+1)}$

$\Rightarrow\:1=\large\frac{5}{r+1}$

$\Rightarrow\:r+1=5\:\:or\:\:r=4$

$i.e.,$ Common ratio $=4$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...