Prove or Show the sum

Question

Let $a_{n}$ be the $n^{\text{th}}$ term of a geometric progression of positive numbers. Let $$\sum_{n=1}^{100}a_{2n}=\alpha$$ and $$\sum_{n=1}^{100}a_{2n-1}=\beta$$ such that $\alpha \ne\beta$. Show that the common ratio is $\frac\alpha\beta$.

Answer 1

For some first term $a$ and some common ratio $r$, the first sum is given by $$a + ar^2 + ar^4+... = a(1+r^2+r^4+...)=\beta.$$ and the second sum is given by $$ar + ar^3 + ar^5+... = a(r+r^3+r^5+...)=\alpha.$$ Dividing these equations, we simply get that $r = \alpha/ \beta$.