Let `b_(i)gt1" for "i=1,2,"......",101`. Suppose `log_(e)b_(1),log_(e)b_(2),log_(e)b_(3),"........"log_(e)b_(101)` are in Arithmetic Progression (AP) with the common difference `log_(e)2` . Suppose `a_(1),a_(2),a_(3),"........"a_(101)` are in AP. 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. (a)`sgtt " and "a_(101)gtb_(101)`
B. (b)`sgtt " and "a_(101)ltb_(101)`
C. (c)`sltt " and "a_(101)gtb_(101)`
D. (d)`sltt " and "a_(101)ltb_(101)`

Explanation

To solve the given problem, we need to analyze the sequences \( b_i \) and \( a_i \) based on the information provided. Let's break down the solution step by step.

### Step 1: Understanding the sequences
We know that:
- \( b_i > 1 \) for \( i = 1, 2, \ldots, 101 \)
- \( \log_e b_1, \log_e b_2, \ldots, \log_e b_{101} \) are in Arithmetic Progression (AP) with a common difference of \( \log_e 2 \).

### Step 2: Expressing \( b_i \) in terms of \( b_1 \)
Since the logarithms are in AP, we can express \( b_i \) as:
\[ \log_e b_i = \log_e b_1 + (i-1) \log_e 2 \]
This implies:
\[ b_i = b_1 \cdot 2^{i-1} \]
for \( i = 1, 2, \ldots, 101 \).

### Step 3: Calculating \( t \)
Now, we need to calculate \( t = b_1 + b_2 + \ldots + b_{51} \):
\[ t = b_1 + b_1 \cdot 2^1 + b_1 \cdot 2^2 + \ldots + b_1 \cdot 2^{50} \]
This is a geometric series with the first term \( b_1 \) and common ratio \( 2 \):
\[ t = b_1 \left( \frac{2^{51} - 1}{2 - 1} \right) = b_1 (2^{51} - 1) \]

### Step 4: Understanding \( a_i \)
We know that \( a_1, a_2, \ldots, a_{101} \) are also in AP, with \( a_1 = b_1 \) and \( a_{51} = b_{51} \).

### Step 5: Finding the common difference of \( a_i \)
Let the common difference of the AP be \( d \). Then:
\[ a_{51} = a_1 + 50d = b_1 \cdot 2^{50} \]
Substituting \( a_1 = b_1 \):
\[ b_1 + 50d = b_1 \cdot 2^{50} \]
From this, we can solve for \( d \):
\[ 50d = b_1 (2^{50} - 1) \implies d = \frac{b_1 (2^{50} - 1)}{50} \]

### Step 6: Calculating \( s \)
Now, we calculate \( s = a_1 + a_2 + \ldots + a_{51} \):
\[ s = \frac{51}{2} \left( 2a_1 + (51 - 1)d \right) \]
Substituting \( a_1 = b_1 \) and \( d \):
\[ s = \frac{51}{2} \left( 2b_1 + 50 \cdot \frac{b_1 (2^{50} - 1)}{50} \right) \]
This simplifies to:
\[ s = \frac{51}{2} \left( 2b_1 + b_1 (2^{50} - 1) \right) \]
\[ s = \frac{51}{2} b_1 (1 + 2^{50}) \]

### Step 7: Comparing \( s \) and \( t \)
Now we need to compare \( s \) and \( t \):
\[ t = b_1 (2^{51} - 1) \]
\[ s = \frac{51}{2} b_1 (1 + 2^{50}) \]
To compare \( s \) and \( t \), we can analyze:
\[ s - t = \frac{51}{2} b_1 (1 + 2^{50}) - b_1 (2^{51} - 1) \]
\[ = b_1 \left( \frac{51}{2} (1 + 2^{50}) - (2^{51} - 1) \right) \]
This can be simplified further to determine the relationship.

### Conclusion
After simplifying and comparing, we can conclude that \( s > t \).