Let a1,a2,a3,…,a10 be in G.P. with a51=25 and ∑i=1101 ai=125. Then the value of ∑i=1101 1ai is

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Detailed Solution

Let the 1st term be a and common ratio be r . Then a1−r1011−r=125∑r=1101 1ai=1a1+1a2+…+1a101 =1a1−r1011−r⋅1r100=125ar502=125625=15

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