Let `b_i > 1` for i =1, 2,....,101. Suppose `log_e b_1, log_e b_2,....,log_e b_101` are in Arithmetic Progression (A.P.) with the common difference `log_e 2`. Suppose `a_1, a_2,...,a_101` are in A.P. such that `a_1 = b_1 and a_51 = b_51`. If `t = b_1 + b_2+.....+b_51 and s = a_1+a_2+....+a_51` then

Options

A. `s gt t and a_(101) gt b_(101)`
B. `s gt t and a_(101) lt b_(101)`
C. `s lt t and a_(101) gt b_(101) gt b_(101)`
D. `s lt t and a_(101) lt b_(101)`

Explanation

To solve the problem step by step, we will analyze the given information and derive the required results.

### Step 1: Understand the Given Information
We have:
- \( b_i > 1 \) for \( i = 1, 2, \ldots, 101 \)
- \( \log_e b_1, \log_e b_2, \ldots, \log_e b_{101} \) are in Arithmetic Progression (A.P.) with a common difference of \( \log_e 2 \).

### Step 2: Express \( b_i \) in terms of \( b_1 \)
Since the logarithms are in A.P., we can express:
\[ \log_e b_i = \log_e b_1 + (i-1) \log_e 2 \]
This implies:
\[ b_i = b_1 \cdot 2^{i-1} \]

### Step 3: Calculate \( t \)
We need to find \( t = b_1 + b_2 + \ldots + b_{51} \):
\[ t = b_1 + b_1 \cdot 2^1 + b_1 \cdot 2^2 + \ldots + b_1 \cdot 2^{50} \]
This is a geometric series with the first term \( b_1 \) and common ratio \( 2 \):
\[ t = b_1 \left( 1 + 2 + 2^2 + \ldots + 2^{50} \right) \]
The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Here, \( a = 1 \), \( r = 2 \), and \( n = 51 \):
\[ t = b_1 \cdot \frac{2^{51} - 1}{2 - 1} = b_1 (2^{51} - 1) \]

### Step 4: Express \( a_i \) in terms of \( a_1 \)
Given that \( a_1 = b_1 \) and \( a_{51} = b_{51} \), we can write:
\[ a_{51} = a_1 + 50d = b_{51} \]
Since \( b_{51} = b_1 \cdot 2^{50} \), we have:
\[ b_1 + 50d = b_1 \cdot 2^{50} \]
This gives us:
\[ 50d = b_1 (2^{50} - 1) \implies d = \frac{b_1 (2^{50} - 1)}{50} \]

### Step 5: Calculate \( s \)
Now, we find \( s = a_1 + a_2 + \ldots + a_{51} \):
\[ s = \frac{51}{2} (a_1 + a_{51}) = \frac{51}{2} (b_1 + b_1 \cdot 2^{50}) = \frac{51}{2} b_1 (1 + 2^{50}) \]

### Step 6: Compare \( s \) and \( t \)
Now we need to find \( s - t \):
\[ s - t = \frac{51}{2} b_1 (1 + 2^{50}) - b_1 (2^{51} - 1) \]
Simplifying this:
\[ s - t = \frac{51}{2} b_1 (1 + 2^{50}) - b_1 (2^{51} - 1) \]
\[ = \frac{51}{2} b_1 + \frac{51}{2} b_1 \cdot 2^{50} - b_1 \cdot 2^{51} + b_1 \]
\[ = \frac{51}{2} b_1 + b_1 - b_1 \cdot 2^{51} + \frac{51}{2} b_1 \cdot 2^{50} \]
Combining terms:
\[ = \left( \frac{53}{2} b_1 + \frac{51}{2} b_1 \cdot 2^{50} - b_1 \cdot 2^{51} \right) \]
\[ = b_1 \left( \frac{53}{2} + \frac{51}{2} \cdot 2^{50} - 2^{51} \right) \]

### Step 7: Conclusion
Since \( b_1 > 1 \) and the expression inside the parentheses is positive, we conclude that \( s > t \).

### Final Result
Thus, we have:
- \( s > t \)
- \( b_{101} > a_{101} \)