Question

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio of the G.P.

Answer 1

Let the G.P. be $2n,2,2n+4,$.....

Then, $S_n=\frac{a(r^n-1)}{r-1},a=2n,r=2$

$\Rightarrow S_n=\frac{2n(2^n-1)}{2-1}=2n^{n+1}-2n$

Then the G.P. of odd term

$a_1+a_3+a_5+.....a_{2n-1}$

According to the question

Sum of all terms = 5 (sum of terms occupying the odd places)

$a_1+a_2+a_3+.....+a_{2n}=5(a_1+a_3+a_5+....a_{2n-1})$

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

$\frac{a(1-r^{2n})}{1-r}=5\left(\frac{a(1-(r{)}^2{)}^n}{1-r^2}\right)$

Take $\left(\frac{a}{1-r}\right)$ common from both sides

$\Rightarrow 1-r^{2n}=\frac{5(1-r^{2n})}{1+r}$

$\Rightarrow 1+r-r^{2n}-r^{2n+1}=5-5r^{2n}$

$\Rightarrow r^{2n+1}-4r^{2n}-r+4=0$

$\Rightarrow r^{2n}(r-4)-1(r-4)=0$

$\Rightarrow (r^{2n}-1)(r-4)=0$

either $r^{2n}=1,r=4$

$\Rightarrow r=4$