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$Let a_{1}, a_{2}, a_{3}, \dots a_{101} are in G.P with a_{51} = 25, and \sum_{i = 1}^{101} a_{i} = 125 . Then the value of \sum_{i = 1}^{101} (\frac{1}{ai}) is$

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Correct option is B

∑i=1101 ai=a1−r1011−r=125Now ∑i=1101 1ai=1a1+1a2+…+1a101 =1a1−r1011−r1r100 =125ar502=125625=1/5

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Let a1,a2,a3,…a101 are in G.P with a51=25, and ∑i=1101 ai=125. Then the value of ∑i=1101 1ai is

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$Let a_{1}, a_{2}, a_{3}, \dots a_{101} are in G.P with a_{51} = 25, and \sum_{i = 1}^{101} a_{i} = 125 . Then the value of \sum_{i = 1}^{101} (\frac{1}{ai}) is$