Answer:
Option B
Explanation:
If $ \log b_{1}, \log b_{2}......... \log b_{101}$ are in AP, with common difference $\log_{e} 2$. then $ b_{1}, b_{2}......... b_{101}$ are in GP , with common ratio 2
$ b_{1}=2^{0}b_{1}$
$ b_{2}=2^{1}b_{1}$
$ b_{3}=2^{2}b_{1}$
: : :
$ b_{101}=2^{100}b_{1}$ ..........(i)
Also $a_{1} ,a_{2},........a_{101}$ are in AP
Given, $a_{1} =b_{1}$ and $a_{51} =b_{51}$
$\Rightarrow a_{1}+50D=2^{50}b_{1}$
$\Rightarrow a_{1}+50D=2^{50}a_{1}[\because a_{1}=b_{1}]$ ........(ii)
Now, $t= b_{1} +b_{2}+.....+b_{51}$
$\Rightarrow t=b_{1}\frac{(2^{51}-1)}{2-1}$........(iii)
and $s= a_{1} +a_{2}+.....+a_{51}$
$=\frac{51}{2}(2a_{1}+50D)$ .....(iv)
$\therefore$ $t= a_1(2^{51}-1)[\because a_{1}=b_{1}]$
or $t= 2^{51}a_{1}-a_{1}<2^{51}a_{1}$ ......(v)
and $s= \frac{51}{2}[a_{1}+(a_{1}+50D]$
[from Eq.(iii)]
= $\frac{51}{2}[a_{1}+2^{50}a_{1}]=\frac{51}{2}a_{1}+\frac{51}{2}2^{50}a_{1}$
$\therefore$ $s>2^{51}a_{1}$ ............(vi)
From Eqs(v) and (vi)
we get s>t
Also $a_{101}=a_{1}+100D$
and $b_{101}=2^{100}b_{1}$
$\therefore$ $a_{101}=a_{1}+100(\frac{2^{50}a_{1}-a_{1}}{50})$
and $b_{101}=2^{100}a_{1}$
$\Rightarrow a_{101}=a_{1}+2^{51}a_{1}-2a_{1}=2^{51}a_{1}-a_{1}$
$\Rightarrow a_{101}<2^{51}a_{1}$
and $ b_{101}>2^{51}a_{1}\Rightarrow b_{101}>a_{101}$
