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

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

$\mathrm{\alpha }-\sum _{\mathrm{n}=1}^{100} {\mathrm{a}}_{2\mathrm{n}}\Rightarrow \mathrm{\alpha }-{\mathrm{a}}_2+{\mathrm{a}}_4+\dots +{\mathrm{a}}_{200}$

$\begin{array}{l}\Rightarrow \mathrm{\alpha }=\mathrm{ar}+{\mathrm{ar}}^3+\dots +{\mathrm{ar}}^{199}\\ \Rightarrow \mathrm{\alpha }=\mathrm{ar}\left(1+{\mathrm{r}}^2+{\mathrm{r}}^4+\dots +{\mathrm{r}}^{198}\right) \left(1\right)\\ \text{ and }\mathrm{\beta }=\sum _{\mathrm{n}=1}^{100} {\mathrm{a}}_{2\mathrm{n}-1}\Rightarrow \mathrm{\beta }={\mathrm{a}}_1+{\mathrm{a}}_3+\dots +{\mathrm{a}}_{199}\\ \Rightarrow \mathrm{\beta }=\mathrm{a}+{\mathrm{ar}}^2+\dots +{\mathrm{ar}}^{198} \left(2\right)\\ \Rightarrow \mathrm{\beta }=\mathrm{a}\left(1+{\mathrm{r}}^2+\dots +{\mathrm{r}}^{198}\right)\end{array}$

From Eqs (1) and (2), we get

   $\frac{\mathrm{\alpha }}{\mathrm{\beta }}=\mathrm{r}$.