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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1
Let ana_n be the nnth term of a GP of positive numbers.
If ∑n=1100a2n=α\sum_{n=1}^{100} a_{2 n}=\alpha and ∑n=1100a2n−1=β\sum_{n=1}^{100} a_{2 n-1}=\beta, such that α≠β\alpha \neq \beta, then the common ratio is
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Correct option: A
αβ\frac{\alpha}{\beta}
Option A (αβ\frac{\alpha}{\beta}) is correct because the ratio of the sums of even and odd terms in a GP directly gives the common ratio.

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