Let $a_{n}$ be $n$th term of the GP of positive numbers. Let $\sum_{n = 1}^{100}{a_{2n} = \alpha}\$ and $\sum_{n = 1}^{100}{a_{2n} = \beta}$, such that $\alpha
eq \beta,$ then the common ratio is

Question

Let $a_{n}$ be $n$th term of the GP of positive numbers. Let $\sum_{n = 1}^{100}{a_{2n} = \alpha}\$ and $\sum_{n = 1}^{100}{a_{2n} = \beta}$, such that $\alpha 
eq \beta,$ then the common ratio is

Dr. Abubacker · Accepted Answer

(a)
Let $'r^{'}$ be the common ratio,
$	herefore\ \frac{\sum_{n = 1}^{100}a_{2n}}{\sum_{n = 1}^{100}a_{2n - 1}} = \frac{a_{2} + a_{4} + a_{6} + ... + a_{200}}{a_{1} + a_{3} + a_{5} + ... + a_{199}}$
$= \ \frac{a_{1}(r + r^{3} + r^{5} + ... + r^{199})}{a_{1}(1 + r^{2} + r^{4} + ... + r^{198})} = r$
$\Rightarrow \ \frac{\alpha}{\beta} = r$

Q. Let $a_{n}$ be $n$th term of the GP of positive numbers. Let $\sum_{n = 1}^{100}{a_{2n} = \alpha}\$ and $\sum_{n = 1}^{100}{a_{2n} = \beta}$, such that $\alpha \neq \beta,$ then the common ratio is

Explanation

Reviewed by Dr. Abubacker

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