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

If \(\log _e b_1, \log _e b_2 \ldots . \log _e b_{101} \rightarrow A P ; \quad\) difference \(( d )=\log _e 2\)
$\Rightarrow b_1,b_2,b_3\mathrm{.......}{b}_{101}\to GP;r=2$
$\therefore b_1,2b_1,2^2{b}_1\mathrm{.........},2^{100}{b}_1\to GP$
$a_1,a_2,a_3\mathrm{..........}{a}_{101}\to AP$ Let Common Difference = D
Given, $a_1=b_1$ and $a_{51}=b_{51}$
\( \Rightarrow a_1+50 D=2^{50} b_1 \)
\( \therefore a_1+50 D=2^{50} a_1\left(A s b_1=a_1\right) \Rightarrow\) \(D=\frac{2^{50} a_1-a_1}{50} \)
\( t=b_1+b_2+\ldots \ldots b_{51}=b_1+2 b_1\) \(+~2^2 b_1+\ldots \ldots 2^{50} b_1 \Rightarrow t=b_1\left(2^{51}-1\right) ; \)
\( s=a_1+a_2+\ldots \ldots a_{51} \Rightarrow s=\frac{51}{2}(2a_1\) \(+~50 D) \)
\( t=a_1 \cdot 2^{51}-a_1 \Rightarrow t1 \text { \& } a_1=b_1\) so \(a_1>1) \)
\( s=\frac{51}{2}\left(a_1+a_1+50 D\right) \)
\( s=\frac{51}{2}\left(a_1+2^{50} a_1\right) \)
\( s=\frac{51 a_1}{2}+\frac{51}{2} \cdot 2^{50} a_1 \)
\( \Rightarrow s>a_1 \cdot 2^{51} \ldots \ldots \text { (ii) }\)
Clearly $s>t$ (from equation (i) and (ii))
Also $a_{101}=a_1+100D; b_{101}=b_1. 2^{100}$
\(\therefore a_{101}=a_1+100\left(\frac{2^{50} a_1-a_1}{50}\right) ; b_{101}\) \(=2^{100} a_1 \ldots \ldots \text { (ii) }\)
\(a_{101}=a_1+2^{51} a_1-2 a_1 \Rightarrow a_{101}=\) \(2^{51} a_1-a_1 \Rightarrow a_{101}<2^{51} a_1 \ldots \ldots .( iv )\)
Clearly $b_{101}>a_{101}$ (from equation (iii) and (iv))