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$Let a_{n} be the n^{th} term of the G.P of positive numbers. Let \sum_{n = 1}^{100} a_{2 n} = α and \sum_{n = 1}^{100} a_{2 n - 1} = β$ $such that α \neq β, then the common ratio is$

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Correct option is A

Let 'a' be the first term and ' r ' the common ratio of given G.P then α=∑n=1100 a2n α=ar⁡1−r2001−r2 β=a1−r2001−r2⇒αβ=r

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Let an be the nth term of the G.P of positive numbers. Let ∑n=1100 a2n=α and ∑n=1100 a2n−1=β such that α≠β , then the common ratio is

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$Let a_{n} be the n^{th} term of the G.P of positive numbers. Let \sum_{n = 1}^{100} a_{2 n} = α and \sum_{n = 1}^{100} a_{2 n - 1} = β$ $such that α \neq β, then the common ratio is$