Let a_n be the nth term of a G.P. of positive numbers. Let $\sum _{\mathrm{n}=1}^{100} {\mathrm{a}}_{2\mathrm{n}}=\mathrm{\alpha }\text{ and }\sum _{\mathrm{n}=1}^{100} {\mathrm{a}}_{2\mathrm{n}-1}=\mathrm{\beta }$, such that $\mathrm{\alpha }\ne \mathrm{\beta }$, then the common ratio is

Let a be the first term and r be the common ratio of the given G.P. Then,

$\begin{array}{c}\mathrm{\alpha }=\sum _{\mathrm{n}=1}^{100} {\mathrm{a}}_{2\mathrm{n}}\\ \Rightarrow \mathrm{\alpha }={\mathrm{a}}_2+{\mathrm{a}}_4+\cdots +{\mathrm{a}}_{200}\\ =\mathrm{ar}+{\mathrm{ar}}^3+\cdots +{\mathrm{ar}}^{199}\\ =\mathrm{ar}\left(1+{\mathrm{r}}^2+{\mathrm{r}}^4+\cdots +{\mathrm{r}}^{198}\right)\\ \mathrm{\beta }=\sum _{\mathrm{n}=1}^{100} {\mathrm{a}}_{2\mathrm{n}-1}\\ \Rightarrow \mathrm{\beta }={\mathrm{a}}_1+{\mathrm{a}}_3+\cdots +{\mathrm{a}}_{199}\\ =\mathrm{a}+{\mathrm{ar}}^2+\cdots +{\mathrm{ar}}^{198}\\ =\mathrm{a}\left(1+{\mathrm{r}}^2+\cdots +{\mathrm{r}}^{198}\right)\end{array}$

Clearly, $\mathrm{\alpha }/\mathrm{\beta }=\mathrm{r}$.