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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1
Let bi>1b_i>1 for i=1,2,…,101i=1,2, \ldots, 101. Suppose log⁡eb1\log _e b_1, log⁡eb2,…,log⁡eb101\log _e b_2, \ldots, \log _e b_{101} are in Arithmetic Progression (A.P) with the common differencelog⁡e2{\log _e}2 . Suppose a1,a2,…,a101a_1, a_2, \ldots, a_{101} are in A.P. such that a1=b1a_1=b_1 and a51=b51a_{51}=b_{51}. If t=b1+b2+…+b51t=b_1+b_2+\ldots+b_{51} and s=a1+a2+…+a51s=a_1+a_2+\ldots+a_{51}, then
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Correct option: B
s>ts>t and a101<b101a_{101} < b_{101}
Option B (s>ts>t and a101<b101a_{101} < b_{101}) is correct because the sum of the arithmetic progression is greater than the sum of the geometric progression, while the 101st term of the arithmetic progression is less than the 101st term of the geometric progression.

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