Let $b_i>1$ for $i=1,2,\dots ,101$ Suppose ${\log }_e{b}_1$,${\log }_e{b}_2,\dots ,{\log }_e{b}_{101}$ 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 

It is given that ${\log }_e{b}_1,{\log }_e{b}_2,\dots ,{\log }_e{b}_{101}$ are in A.P. with common difference ${\log }_{e}2$. Therefore, $b_1,b_2,b_3,\dots ,b_{101}$ are in G.P. with common ratio 2.

Let $d$ be the common difference of the A.P. $a_1,a_2,a_2,a_{101}$.

Now, $a_{51}=b_{51}$

$\Rightarrow a_1+50d=a_1\left(2^{50}\right) \left[\because b_1=a_1\right]$

$\Rightarrow 50d=\left(2^{50}-1\right)a_1$                             ..(i)

  $\therefore a_{101}=a_1+100d=a_1+2\left(2^{50}-1\right)a_1=\left(2^{51}-1\right)a_1$

and, $b_{101}=b_1\times 2^{100}=a_1\left(2^{100}\right)$

Clearly, $b_{101}>a_{101}$

Now,

      $\begin{array}{l}s=\frac{51}{2}\left(2a_1+50d\right)=51a_1+\frac{50\times 51d}{2}\\ \Rightarrow s=51a_1+\frac{51}{2}\left(2^{50}\mid -1\right)a_1\end{array}$                               [Using (i)]

and, $t=b_1+b_2+\dots +b_{51}$

$\begin{array}{l}\Rightarrow t=b_1\frac{\left(2^{51}-1\right)}{2-1}=a_1\left(2^{51}-1\right)\\ \therefore s-t=\frac{51a}{2}\left(2^{50}+1\right)-a_1\left(2^{51}-1\right)=\left(47\times 2^{49}+\frac{53}{2}\right)a_1>0\\ \Rightarrow s>t\end{array}$

Hence, $s>t\text{ and }{a}_{101}<b_{101}$.

So, option (b) is correct.