Question

If an be the $n^{\text{th }}$ term of the GP of positive numbers. Let $\sum _{n=100}^{100}{a}_{2n}=\alpha$ and $\sum _{n=1}^{100}{a}_{2n-1}=\beta$ such that $\alpha \ne \beta$ then the common ratio is:

• (A) $\frac{\alpha }{\beta }$
• (B) $\frac{\beta }{\alpha }$
• (C) $\sqrt{\frac{\alpha }{\beta }}$
• (D) $\sqrt{\frac{\beta }{\alpha }}$

Answer 1

The Correct Option is A

Explanation:
Given:A geometric progression such that $n^{\text{th }}$ term is $a_n$ and$\sum _{n=1}^{100}{a}_{2n}=\alpha ,\sum _{n=1}^{100}{a}_{2n-1}=\beta$We have to find the common ratio of given GP.Let a be the first term, r be the common ratio of the given GP $(r>0)$.Then $\mathrm{a},\mathrm{ar},{\mathrm{ar}}^2,{\mathrm{ar}}^3,\dots .,{\mathrm{ar}}^{199}$ are in GP.We shall use the formula of general term of GP.$\alpha =\sum _{n=1}^{100}{a}_{2n}\Rightarrow a=a_2+a_4+\dots \dots .+a_{200}$$\Rightarrow a=ar+a{r}^3+\dots \dots ..+a{r}^{199}=\mathrm{ar}(1+r^2+\dots \dots +r^{198})$and $\beta =\sum _{n=1}^{100}{a}_{2n-1}\Rightarrow \beta =a_1+a_3+\dots \dots \dots +a_{199}$$\Rightarrow \beta =a+a{r}^2+\dots \dots ..+{\mathrm{ar}}^{198}$$\Rightarrow \beta =a(1+r^2+\dots \dots .+r^{198})$Clearly, $\mathrm{a}/\beta =r=$ common ratioHence, the correct option is (A).