Let $b_i > 1$ for $i = 1, 2, \ldots, 101$. Suppose $\log_e b_1, \log_e b_2, \ldots, \log_e b_{101}$ are in Arithmetic Progression (A.P.) with the common difference $\log_e 2$. Suppose $a_1, a_2, \ldots, a_{101}$ are in A.P. such that $a_1 = b_1$ and $a_{51} = b_{51}$. If $t = b_1 + b_2 + \ldots + b_{51}$ and $s = a_1 + a_2 + \ldots + a_{51}$, then

Let bᵢ > 1 for i = 1, 2, …, 101. Suppose logₑ b₁, logₑ b₂, …, logₑ b₁₀₁ are in Arithmetic Progression (A.P.) with the common difference logₑ 2. Suppose a₁, a₂, …, a₁₀₁ are in A.P. such that a₁ = b₁ and a₅₁ = b₅₁. If t = b₁ + b₂ + … + b₅₁ and s = a₁ + a₂ + … + a₅₁, then

$b_i > 1,\quad i = 1, 2, \ldots, 101$

$\log_e b_1, \log_e b_2, \ldots, \log_e b_{101} \ \text{in A.P.}$

$\text{common difference } d = \log_e 2$

$a_1 = b_1,\qquad a_{51} = b_{51}$

$t = b_1 + b_2 + \ldots + b_{51}$

$s = a_1 + a_2 + \ldots + a_{51}$

$\log_e b_i = \log_e b_1 + (i-1)\log_e 2$

$\Rightarrow\ b_i = b_1 \cdot 2^{\,i-1}$

$\text{so } b_i \text{ is a G.P. with common ratio } 2$

$b_{51} = b_1 \cdot 2^{50}$

$d = \dfrac{a_{51} - a_1}{50} = \dfrac{b_1(2^{50} - 1)}{50}$

$a_i = a_1 + (i-1)d$

$a_i = b_1 + (i-1)\,\dfrac{b_1(2^{50} - 1)}{50}$

$\text{the chord } a_i \text{ lies above the convex } b_i$

$a_i \ge b_i \quad (1 < i < 51)$

$s - t = \sum_{i=1}^{51}(a_i - b_i) > 0$

$\therefore\ s > t$

$t = b_1 + b_1 2 + \ldots + b_1 2^{50}$

$t = b_1\,\dfrac{2^{51}-1}{2-1} = b_1(2^{51}-1)$

$s = \dfrac{51}{2}(a_1 + a_{51}) = \dfrac{51}{2}\,b_1(1 + 2^{50})$

$s - t = b_1\!\left(\dfrac{53}{2} + \dfrac{47}{2}\cdot 2^{50}\right) > 0$

$a_{101} = a_1 + 100\,d$

$= b_1 + 100\cdot\dfrac{b_1(2^{50}-1)}{50}$

$= b_1(2^{51}-1)$

$b_{101} = b_1 \cdot 2^{100}$

$a_{101} = b_1(2^{51}-1) < b_1\cdot 2^{100} = b_{101}$

$\therefore\ a_{101} < b_{101}$

$s > t \quad \text{and} \quad a_{101} < b_{101}$

$\boxed{\text{Answer: (2)}}$