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

(c) 4

$$\begin{gathered} \mathrm{Let}\mathrm{there}\mathrm{be}2n\mathrm{terms}\mathrm{in}\mathrm{a}\mathrm{G}.\mathrm{P}. \\ \mathrm{Let}a\mathrm{be}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}r\mathrm{be}\mathrm{the}\mathrm{common}\mathrm{ratio}. \\ \because S_{2n}=5\left(S_{\mathrm{odd}\mathrm{terms}}\right) \\ \Rightarrow \frac{a\left(r^{2n}-1\right)}{\left(r-1\right)}=5\left(a+a{r}^2+a{r}^4+a{r}^6+...a{r}^{\left(2n-1\right)}\right) \\ \Rightarrow \frac{a\left(r^{2n}-1\right)}{\left(r-1\right)}=5\left(\frac{a\left({\left(r^2\right)}^n-1\right)}{\left(r^2-1\right)}\right) \\ \Rightarrow \frac{\left(r^{2n}-1\right)}{\left(r-1\right)}=5\frac{\left({\left(r^2\right)}^n-1\right)}{\left(r^2-1\right)} \\ \Rightarrow \frac{\left({\left(r^n\right)}^2-1^2\right)}{\left(r-1\right)}=5\frac{\left({\left(r^n\right)}^2-1^2\right)}{\left(r^2-1\right)} \\ \Rightarrow \frac{\left(r^n-1\right)\left(r^n+1\right)}{\left(r-1\right)}=5\frac{\left(r^n-1\right)\left(r^n+1\right)}{\left(r-1\right)\left(r+1\right)} \\ \Rightarrow \left(r^n-1\right)\left(r^n+1\right)\left(r-1\right)\left(r+1\right)-5\left(r-1\right)\left(r^n-1\right)\left(r^n+1\right)=0 \\ \Rightarrow \left(r^n-1\right)\left(r^n+1\right)\left(r-1\right)\left(r+1-5\right)=0 \\ \mathrm{But},r=1\mathrm{or}-1\mathrm{is}\mathrm{not}\mathrm{possible}. \\ \therefore r=4 \end{gathered}$$