$Let b_{i} > 1 for i = 1, 2, \dots, 101 . Suppose \log_{ε} b_{1}, \log_{ε} b_{2}, \dots, \log_{ε} b_{101} are in arithmetic progression$
$(A.P). With the common difference \log_{e} 2 . Suppose a_{1}, a_{2}, \dots ., a_{101} are in A.P such that$ $a_{1} = b_{1} and a_{51} = b_{51} .If t = b_{1} + b_{2} + \dots . . + b_{51} and s = a_{1} + a_{2} + \dots . + a_{51 .} then$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$s > t a n d a_{101} > b_{101}$

$s > t a n d a_{101} < b_{101}$

$s < t a n d a_{101} > b_{101}$

$s < t a n d a_{101} < b_{101}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} t = b_{1} (1 + 2 + 2^{2} + {....2}^{50}) = a_{1} (2^{51} - 1) \\ s = \frac{51}{2} (a_{1} + a_{51}) = \frac{51}{2} a_{1} (1 + 2^{50}) \\ \Rightarrow t < r \\ \Rightarrow a_{101} = a_{1} + 100 d = a_{1} + 2 (2^{50} - 1) = a_{1} (2^{51} - 1) \\ \Rightarrow b_{101} = a_{1} 2^{100} \end{array}$