Let an be the the nth term of a G.P. of positive numbers such that 100∑n=1a2n=α and 100∑n=1a2n−1=β,α≠β. Then the common ratio of G.P. is
A
αβ
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B
βα
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C
√αβ
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D
√βα
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Solution
The correct option is Aαβ Let first term and common ratio of G.P. be a and r respectively. 100∑n=1a2n=α ⇒a2+a4+a6+⋯+a200=α ⇒ar+ar3+ar5+⋯+ar199=α ⇒ar(1+r2+r4+⋯+r198)=α⋯(1)
100∑n=1a2n−1=β ⇒a1+a3+a5+⋯+a199=β ⇒a+ar2+⋯+ar198=β ⇒a(1+r2+⋯+r198)=β⋯(2) Dividing (1) and (2), we get r=αβ