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GeneralMedium JEE math MCQ

Let $b_{i}>1$ for $i=1,2, \ldots, 101$. Suppose $\log _{e} b_{1}, \log _{e} b_{2}, \ldots, \log _{e} b_{101}$ are in Arithmetic Progression (A.P.) with the common difference $\log _{e} 2$. Suppose $a_{1}, a_{2}, \ldots, a_{101}$ are in A.P. such that $a_{1}=b_{1}$ and $a_{51}=b_{51}$. If $t=b_{1}+b_{2}+\cdots+b_{51}$ and $s=a_{1}+a_{2}+\cdots+a_{51}$, then
  1. A. $s>t$ and $a_{101}>b_{101}$
  2. B. $s>t$ and $a_{101}<b_{101}$
  3. C. $s<t$ and $a_{101}>b_{101}$
  4. D. $s<t$ and $a_{101}<b_{101}$

Solution

The correct option is **B**.

MATH

mediumPYQ Reworded
Question
Read carefully, then pick the best option.
Let bi>1b_{i}>1 for i=1,2,,101i=1,2, \ldots, 101. Suppose logeb1,logeb2,,logeb101\log _{e} b_{1}, \log _{e} b_{2}, \ldots, \log _{e} b_{101} are in Arithmetic Progression (A.P.) with the common difference loge2\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}+\cdots+b_{51} and s=a1+a2++a51s=a_{1}+a_{2}+\cdots+a_{51}, then
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