Let `X_(n)` denote the mean of first n natural numbers, then the mean of `X_(1), X_(2), ………….., X_(100)` is

Options

A. 25
B. 50
C. 25.5
D. 25.75

Answer

To solve the problem, we need to find the mean of the values \( X_1, X_2, \ldots, X_{100} \), where \( X_n \) is defined as the mean of the first \( n \) natural numbers.

### Step-by-Step Solution:

1. **Define \( X_n \)**:
The mean of the first \( n \) natural numbers is given by:
\[ X_n = \frac{\text{Sum of first } n \text{ natural numbers}}{n} \]
The sum of the first \( n \) natural numbers is:
\[ S_n = \frac{n(n + 1)}{2} \]
Therefore, we can express \( X_n \) as:
\[ X_n = \frac{S_n}{n} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

2. **Calculate \( X_1, X_2, \ldots, X_{100} \)**:
Now, we can calculate the values of \( X_1, X_2, \ldots, X_{100} \):
- \( X_1 = \frac{1 + 1}{2} = 1 \)
- \( X_2 = \frac{2 + 1}{2} = \frac{3}{2} \)
- \( X_3 = \frac{3 + 1}{2} = 2 \)
- Continuing this way, we find:
- \( X_n = \frac{n + 1}{2} \)
- Thus, \( X_{100} = \frac{100 + 1}{2} = \frac{101}{2} \)

3. **Sum of \( X_1, X_2, \ldots, X_{100} \)**:
We need to find the sum:
\[ \sum_{n=1}^{100} X_n = \sum_{n=1}^{100} \frac{n + 1}{2} \]
This can be simplified as:
\[ = \frac{1}{2} \sum_{n=1}^{100} (n + 1) = \frac{1}{2} \left( \sum_{n=1}^{100} n + \sum_{n=1}^{100} 1 \right) \]
The first sum is the sum of the first 100 natural numbers:
\[ \sum_{n=1}^{100} n = \frac{100 \times 101}{2} = 5050 \]
The second sum is simply 100 (since we are adding 1 a total of 100 times):
\[ \sum_{n=1}^{100} 1 = 100 \]
Therefore, we have:
\[ \sum_{n=1}^{100} X_n = \frac{1}{2} (5050 + 100) = \frac{5150}{2} = 2575 \]

4. **Calculate the Mean**:
The mean of \( X_1, X_2, \ldots, X_{100} \) is given by:
\[ \text{Mean} = \frac{\sum_{n=1}^{100} X_n}{100} = \frac{2575}{100} = 25.75 \]

### Final Answer:
The mean of \( X_1, X_2, \ldots, X_{100} \) is \( \boxed{25.75} \).