If logb₁, logb₂, ......, logb₁₀₁ are in A.P. with common difference log_e2, then b₁, b₂, ......, b₁₀₁ are in G.P., with common ratio 2.

$$\therefore$$ b₁ = 2⁰b₁

b₂ = 2¹b₁

b₃ = 2²b₁

$$\matrix{ \vdots & \vdots & \vdots \cr } $$

b₁₀₁ = 2¹⁰⁰b₁ ..... (i)

Also, a₁, a₂, ......, a₁₀₁ are in A.P.

Given, a₁ = b₁ and a₅₁ = b₅₁

$$\Rightarrow$$ a₁ = b₁ and a₅₁ = b₅₁

$$\Rightarrow$$ a₁ + 50D = 2⁵⁰ b₁

$$\Rightarrow$$ a₁ + 50D = 2⁵⁰ a₁ [$$\because$$ a₁ = b₁] ...... (ii)

Now, t = b₁ + b₂ + ..... + b₅₁

$$ \Rightarrow t = {b_1}{{({2^{51}} - 1)} \over {2 - 1}}$$ ..... (iii)

and s = a₁ + a₂ + .... + a₅₁

$$ = {{51} \over 2}(2{a_1} + 50D)$$ ...... (iv)

$$\therefore$$ t = a₁(2⁵¹ $$-$$ 1) [$$\because$$ a₁ = b₁]

or t = 2⁵¹ a₁ $$-$$ a₁ < 2⁵¹ a₁ ...... (v)

and $$s = {{51} \over 2}[{a_1} + ({a_1} + 50D)]$$ [from Eq. (ii)]

$$ = {{51} \over 2}[{a_1} + {2^{50}}{a_1}] = {{51} \over 2}{a_1} + {{51} \over 2}{2^{50}}{a_1}$$

$$\therefore$$ s > 2⁵¹ a₁ ...... (vi)

From Eqs. (v) and (vi),

we get s > t

Also, a₁₀₁ = a₁ + 100 D

and b₁₀₁ = 2¹⁰⁰ b₁

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

and b₁₀₁ = 2¹⁰⁰ a₁

$$\Rightarrow$$ a₁₀₁ = a₁ + 2⁵¹ a₁ $$-$$ 2a₁ = 2⁵¹ a₁ $$-$$ a₁

$$\Rightarrow$$ a₁₀₁ < 2⁵¹ a₁

and b₁₀₁ > 2⁵¹ a₁ $$\Rightarrow$$ b₁₀₁ > a₁₀₁