Question

In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is

• A. $-\frac{4}{5}$
• B. $\frac{1}{5}$
• C. 4
• D. none of these

Answer 1

The correct option is C

4

(c) 4

Let the be 2n terms in a G.P.

Let a be the first term and r be the common ratio

$S_{2n}=5\left(S_{(}odd\right)terms)$

$\Rightarrow \frac{a(r^{3}n-1)}{(r-1)}$

$=5(a+a{r}^2+a{r}^4+a{r}^6+....a{r}^{(2n-1)})$

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

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

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

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

$(r^n-1)(r^n+1)(r-1)(r+1)-5(r-1)(r^n-1)(r^n+1)=0$

$\Rightarrow (r^n-1)(r^n+1)(r-1)(r+1-5)=0$

But, r = 1 or - 1 is not possible.

$\therefore$ r = 4