Let {a_n} be the {n^{th}} term of the G.P. of | vedclass

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Question

(a) Let the G.P. be $a,\;ar,\;a{r^2}.......,$ then

$\alpha = \sum\limits_{n = 1}^{100} {{a_{2n}}} = {a_2} + {a_4} + .......{\rm{upto}}\;100\;{\rm{t

Vedclass · Accepted Answer

$\frac{\alpha }{\beta }$

Vedclass · Answer

(a) Let the G.P. be $a,\;ar,\;a{r^2}.......,$ then

$\alpha = \sum\limits_{n = 1}^{100} {{a_{2n}}} = {a_2} + {a_4} + .......{\rm{upto}}\;100\;{\rm{terms}}$

$ = ar + a{r^3} + .......{\rm{upto}}\;100\;{\rm{terms}}$

$ = ar(1 + {r^2} + {r^4} + ......{r^{198}})$ and $\beta = \sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = a + a{r^3} + .....{\rm{upto}}\;100\;{\rm{terms}}$

$ = a(1 + {r^2} + ...... + {r^{198}})$

Obviously $\frac{\alpha }{\beta } = r$.

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

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