Question

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

Answer 1

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

$$\begin{gathered} \therefore \sum _{n=1}^{100}{a}_{2n}=\alpha \mathrm{and}\sum _{n=1}^{100}{a}_{2n-1}=\beta \\ \therefore a_2+a_4+...+a_{200}=\alpha \mathrm{and}{a}_1+a_3+...+a_{199}=\beta \\ \Rightarrow ar+a{r}^3+...+a{r}^{199}=\alpha \mathrm{and}a+a{r}^2+...+a{r}^{198}=\beta \\ \Rightarrow ar\left\{\frac{1-{\left(r^2\right)}^{100}}{1-r^2}\right\}=\alpha \mathrm{and}a\left\{\frac{1-{\left(r^2\right)}^{100}}{1-r^2}\right\}=\beta \\ \mathrm{Now},\mathrm{dividing}\alpha \mathrm{by}\beta \\ \frac{\alpha }{\beta }=\frac{ar\left\{\frac{1-{\left(r^2\right)}^{100}}{1-r^2}\right\}}{a\left\{\frac{1-{\left(r^2\right)}^{100}}{1-r^2}\right\}}=\frac{ar}{r}=r \\ \therefore r=\frac{\alpha }{\beta } \end{gathered}$$