Calculate the mean deviation about the mean of the set of first `n` natural numbers when `n` is even natural number.

Explanation

To calculate the mean deviation about the mean of the set of the first `n` natural numbers when `n` is an even natural number, we can follow these steps:

### Step 1: Define `n`
Since `n` is an even natural number, we can express it as:
\[ n = 2m \]
where \( m \) is a natural number.

### Step 2: List the first `n` natural numbers
The first `n` natural numbers are:
\[ 1, 2, 3, \ldots, 2m \]

### Step 3: Calculate the mean
The mean \( \bar{x} \) of the first `n` natural numbers is calculated using the formula for the sum of the first `n` natural numbers:
\[ \bar{x} = \frac{1 + 2 + 3 + \ldots + 2m}{2m} = \frac{2m(2m + 1)/2}{2m} = \frac{2m + 1}{2} \]

### Step 4: Calculate the deviations
The deviation \( d_i \) for each number \( x_i \) from the mean \( \bar{x} \) is given by:
\[ d_i = |x_i - \bar{x}| \]
We can split the deviations into two parts:
1. For \( i = 1 \) to \( m \) (where \( x_i < \bar{x} \)):
\[ d_i = \bar{x} - x_i \]
2. For \( i = m + 1 \) to \( 2m \) (where \( x_i > \bar{x} \)):
\[ d_i = x_i - \bar{x} \]

### Step 5: Calculate the total deviation
The total deviation \( D \) can be calculated as:
\[ D = \sum_{i=1}^{m} (\bar{x} - x_i) + \sum_{i=m+1}^{2m} (x_i - \bar{x}) \]
Calculating each part:
- For the first part:
\[ \sum_{i=1}^{m} (\bar{x} - x_i) = m \cdot \bar{x} - \sum_{i=1}^{m} x_i = m \cdot \frac{2m + 1}{2} - \frac{m(m + 1)}{2} \]
- For the second part:
\[ \sum_{i=m+1}^{2m} (x_i - \bar{x}) = \sum_{i=m+1}^{2m} x_i - m \cdot \bar{x} = \left( \frac{(2m)(2m + 1)}{2} - \frac{m(m + 1)}{2} \right) - m \cdot \frac{2m + 1}{2} \]

### Step 6: Combine the results
Combining both parts gives:
\[ D = m \cdot \frac{2m + 1}{2} - \frac{m(m + 1)}{2} + \left( \frac{(2m)(2m + 1)}{2} - \frac{m(m + 1)}{2} \right) - m \cdot \frac{2m + 1}{2} \]
This simplifies to:
\[ D = m^2 \]

### Step 7: Calculate the mean deviation
The mean deviation \( MD \) is given by:
\[ MD = \frac{D}{n} = \frac{m^2}{2m} = \frac{m}{2} \]
Since \( m = \frac{n}{2} \):
\[ MD = \frac{n}{4} \]

### Final Result
Thus, the mean deviation about the mean of the set of the first `n` natural numbers when `n` is an even natural number is:
\[ \frac{n}{4} \]