Compare Sums and Terms of Arithmetic and Geometric Progressions

To determine the correct relationship between $s$ and $t$, and between $a_{101}$ and $b_{101}$, we analyze the properties of the given sequences step-by-step.

Step 1: Analyze the sequence $b_i$

We are given that $\log_e b_1, \log_e b_2, \dots, \log_e b_{101}$ form an Arithmetic Progression (A.P.) with a common difference of $\log_e 2$.

Thus, for any term $b_i$: $\log_e b_i = \log_e b_1 + (i - 1)\log_e 2 = \log_e \left(b_1 \cdot 2^{i-1}\right)$

Taking the exponential on both sides, we get: $b_i = b_1 \cdot 2^{i-1} \quad \text{for } i = 1, 2, \dots, 101$

Hence, $b_1, b_2, \dots, b_{101}$ form a Geometric Progression (G.P.) with first term $b_1 > 1$ and common ratio $r = 2$.

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Step 2: Analyze the sequence $a_i$

The sequence $a_1, a_2, \dots, a_{101}$ forms an A.P. with common difference $d$. We are given: $a_1 = b_1 \quad \text{and} \quad a_{51} = b_{51}$

From the formula for the $n$-th term of an A.P. and G.P.: $a_{51} = a_1 + 50d = b_1 + 50d$ $b_{51} = b_1 \cdot 2^{50}$

Equating $a_{51}$ and $b_{51}$: $b_1 + 50d = b_1 \cdot 2^{50} \implies 50d = b_1(2^{50} - 1) \implies d = \frac{b_1(2^{50} - 1)}{50}$

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Step 3: Compare $s$ and $t$

We are given: $t = \sum_{i=1}^{51} b_i = b_1 + b_2 + \dots + b_{51}$ $s = \sum_{i=1}^{51} a_i = a_1 + a_2 + \dots + a_{51}$

Since $a_i$ is an A.P., its sum $s$ can be written using the first and last terms as: $s = \frac{51}{2}(a_1 + a_{51}) = \frac{51}{2}(b_1 + b_{51})$

Now, consider the function $f(x) = b_1 \cdot 2^{x-1}$ for $x \in [1, 51]$. Since $f''(x) = b_1 (\ln 2)^2 2^{x-1} > 0$, the function $f(x)$ is strictly convex.

By the property of strictly convex functions, a secant line connecting two points $(1, f(1))$ and $(51, f(51))$ lies strictly above the curve $y = f(x)$ for all intermediate points $x \in (1, 51)$. Here, the linear sequence $a_i$ represents the secant line values, so: $a_i > b_i \quad \text{for all } i \in \{2, 3, \dots, 50\}$

Since $a_1 = b_1$ and $a_{51} = b_{51}$, summing all terms from $i = 1$ to $51$ gives: $\sum_{i=1}^{51} a_i > \sum_{i=1}^{51} b_i \implies s > t$

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Step 4: Compare $a_{101}$ and $b_{101}$

Using the value of $d$ derived in Step 2, we calculate $a_{101}$: $a_{101} = a_1 + 100d = b_1 + 2(50d)$ $a_{101} = b_1 + 2 b_1 (2^{50} - 1) = b_1 \left(1 + 2^{51} - 2\right) = b_1 (2^{51} - 1)$

Next, we calculate $b_{101}$: $b_{101} = b_1 \cdot 2^{100}$

Comparing $a_{101}$ and $b_{101}$: $b_{101} - a_{101} = b_1 \cdot 2^{100} - b_1 (2^{51} - 1) = b_1 \left(2^{100} - 2^{51} + 1\right)$

Since $2^{100} > 2^{51}$ and $b_1 > 1$, it is evident that $b_{101} - a_{101} > 0$, which means: $a_{101} < b_{101}$

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Conclusion

Combining both results: $s > t \quad \text{and} \quad a_{101} < b_{101}$

This corresponds to Option B.