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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
$\frac{\alpha }{\beta }$
$\frac{\beta }{\alpha }$
$\sqrt {\frac{\alpha }{\beta }} $
$\sqrt {\frac{\beta }{\alpha }} $
Solution
(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 $\frac{\alpha }{\beta } = r$.
