General — Medium 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
- A. $s>t$ and $a_{101}>b_{101}$
- B. $s>t$ and $a_{101}<b_{101}$
- C. $s<t$ and $a_{101}>b_{101}$
- D. $s<t$ and $a_{101}<b_{101}$
Solution
The correct option is **B**.
