Let b_i > 1 for i = 1, 2, ..., 101. Suppose loge b₁ loge b₂, ..., loge b₁₀₁ are in arithmetic progression (A.P.) with the common difference 1og_e 2. Suppose a₁, a₂, …, a₁₀₁, are in A.P. such that a₁ = b₁ and ${\mathrm{a}}_{51}={\mathrm{b}}_{51}.\text{ If }\mathrm{t}={\mathrm{b}}_1+{\mathrm{b}}_2+\cdots +{\mathrm{b}}_{51}$ and s = a₁ + a₂ + .. . + a₅₁, then

We have $\log {\mathrm{b}}_2-\log {\mathrm{b}}_1=\log \left(2\right)$

$\Rightarrow \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=2$

$\Rightarrow {\mathrm{b}}_1, {\mathrm{b}}_2\dots$ are in are in G.P. with common ratio 2

$\therefore \mathrm{t}={\mathrm{b}}_1+2{\mathrm{b}}_1+\cdots +2^{50}{\mathrm{b}}_1={\mathrm{b}}_1\left(2^{51}-1\right)$

$\begin{array}{c}\mathrm{s}={\mathrm{a}}_1+{\mathrm{a}}_2+\cdots +{\mathrm{a}}_{51}\\ =\frac{51}{2}\left({\mathrm{a}}_1+{\mathrm{a}}_{51}\right)\\ =\frac{51}{2}\left({\mathrm{b}}_1+{\mathrm{b}}_{51}\right)\\ =\frac{51}{2}{\mathrm{b}}_1\left(1+2^{50}\right)\end{array}$

$\begin{array}{c}\Rightarrow \mathrm{s}-\mathrm{t}={\mathrm{b}}_1\left(\frac{51}{2}+51\times 2^{49}-2^{51}+1\right)\\ ={\mathrm{b}}_1\left(\frac{53}{2}+2^{49}\times 47\right)>0\end{array}$

$\Rightarrow \mathrm{s}>\mathrm{t}$

$\begin{array}{c}{\mathrm{b}}_{101}=2^{100}{\mathrm{b}}_1\\ {\mathrm{a}}_{101}={\mathrm{a}}_1+100\mathrm{d}\\ =2\left({\mathrm{a}}_1+50\mathrm{d}\right)-{\mathrm{a}}_1\\ =2{\mathrm{a}}_{51}-{\mathrm{a}}_1\\ =2{\mathrm{b}}_{51}-{\mathrm{b}}_1\\ =\left(2\times 2^{51}-1\right){\mathrm{b}}_1\\ =\left(2^{51}-1\right){\mathrm{b}}_1\end{array}$

$\therefore {\mathrm{b}}_{101}>{\mathrm{a}}_{101}$