Question

Let a_n be the nth term of the G.P. of positive numbers.

Let \[\sum^{100}_{n = 1} a_{2n} = \alpha \text { and } \sum^{100}_{n = 1} a_{2n - 1} = \beta,\] 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.

\[\therefore \sum^{100}_{n = 1} a_{2n} = \alpha \text { and } \sum^{100}_{n = 1} a_{2n - 1} = \beta\]

\[ \therefore a_2 + a_4 + . . . + a_{200} = \alpha \text { and } a_1 + a_3 + . . . + a_{199} = \beta\]

\[ \Rightarrow ar + a r^3 + . . . + a r^{199} = \alpha \text { and } a + a r^2 + . . . + a r^{198} = \beta\]

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

\[\text { Now, dividing } \alpha \text { by }\beta\]

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

\[ \therefore r = \frac{\alpha}{\beta}\]