# Solve the following

Question:

Let $a_{n}$ be the $n$th term of the G.P. of positive numbers. Let $\sum_{n=1}^{100} a_{2 n}=\alpha$ and $\sum_{n=1}^{100} a_{2 n-1}=\beta$, such that $\alpha \neq \beta$. Prove that the common ratio of the G.P. is $\alpha / \beta$.

Solution:

Let be the first term and r be the common ratio of the G.P.

$\therefore \sum_{n=1}^{100} a_{2 n}=\alpha$ and $\sum_{n=1}^{100} a_{2 n-1}=\beta$

$\therefore a_{2}+a_{4}+\ldots+a_{200}=\alpha$ and $a_{1}+a_{3}+\ldots+a_{199}=\beta$

$\Rightarrow a r+a r^{3}+\ldots+a r^{199}=\alpha$ and $a+a r^{2}+\ldots+a r^{198}=\beta$

$\Rightarrow a r\left\{\frac{1-\left(r^{2}\right)^{100}}{1-r^{2}}\right\}=\alpha$ and $a\left\{\frac{1-\left(r^{2}\right)^{100}}{1-r^{2}}\right\}=\beta$

Now, dividing $\alpha$ by $\beta$

$\frac{\alpha}{\beta}=\frac{a r\left\{\frac{1-\left(r^{2}\right)^{100}}{1-r^{2}}\right\}}{a\left\{\frac{1-\left(r^{2}\right)^{100}}{1-r^{2}}\right\}}=\frac{a r}{r}=r$

$\therefore r=\frac{\alpha}{\beta}$