Question

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

Answer 1

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

$\therefore {\sum }_{n=1}^{100}{a}_{2n}=\alpha$ and ${\sum }_{n=1}^{100}{a}_{2n-1}=\beta$

$\therefore a^2+a_4+....+a_{200}=\alpha and{a}_1+a_3+....a_{199}=\beta$

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

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

Now dividing $\alpha and\beta$

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

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