Mean deviations of the series `a,a+d,a+2d, . . .,a+2nd` from its mean is

Options

A. `(n(n+1)d)/((2n+1))`
B. `(nd)/(2n+1)`
C. `((n+1)d)/(2n+1)`
D. `((2n+1)d)/(n(n+1))`

Answer

To find the mean deviation of the series \( a, a+d, a+2d, \ldots, a+2nd \) from its mean, we will follow these steps:

### Step 1: Identify the Series
The series is an arithmetic progression (AP) given by:
\[ a, a+d, a+2d, \ldots, a+2nd \]

### Step 2: Determine the Number of Terms
The number of terms in the series can be calculated as follows:
- The first term is \( a \) (when \( n=0 \)).
- The last term is \( a + 2nd \) (when \( n=2n \)).
- The total number of terms \( N \) is \( 2n + 1 \).

### Step 3: Calculate the Mean
The mean \( \bar{x} \) of the series can be calculated using the formula for the mean of an arithmetic series:
\[ \bar{x} = \frac{\text{Sum of all terms}}{\text{Number of terms}} \]
The sum of the series can be calculated as:
\[ \text{Sum} = a + (a+d) + (a+2d) + \ldots + (a+2nd) \]
This can be simplified:
\[ \text{Sum} = (2n + 1)a + d(0 + 1 + 2 + \ldots + 2n) \]
Using the formula for the sum of the first \( m \) natural numbers, \( 0 + 1 + 2 + \ldots + m = \frac{m(m+1)}{2} \), we have:
\[ 0 + 1 + 2 + \ldots + 2n = \frac{2n(2n+1)}{2} = n(2n+1) \]
Thus, the sum becomes:
\[ \text{Sum} = (2n + 1)a + d \cdot n(2n + 1) = (2n + 1)(a + nd) \]
Now, substituting back into the mean formula:
\[ \bar{x} = \frac{(2n + 1)(a + nd)}{2n + 1} = a + nd \]

### Step 4: Calculate the Mean Deviation
The mean deviation \( D \) is defined as:
\[ D = \frac{\sum |x_i - \bar{x}|}{N} \]
Where \( x_i \) are the terms of the series. We will calculate \( |x_i - \bar{x}| \) for each term:
\[ |x_i - \bar{x}| = |(a + kd) - (a + nd)| = |kd - nd| = |(k - n)d| \]
Where \( k \) ranges from \( 0 \) to \( 2n \).

Now, we calculate the sum of the absolute deviations:
\[ \sum |x_i - \bar{x}| = \sum_{k=0}^{2n} |(k - n)d| = d \sum_{k=0}^{2n} |k - n| \]
This can be split into two parts:
- For \( k = 0 \) to \( n \): \( n - k \)
- For \( k = n+1 \) to \( 2n \): \( k - n \)

Calculating these sums:
1. From \( k = 0 \) to \( n \):
\[ \sum_{k=0}^{n} (n - k) = n(n + 1)/2 \]
2. From \( k = n+1 \) to \( 2n \):
\[ \sum_{k=n+1}^{2n} (k - n) = \sum_{j=1}^{n} j = n(n + 1)/2 \]
Thus, the total sum of absolute deviations is:
\[ \sum |x_i - \bar{x}| = d \left( \frac{n(n + 1)}{2} + \frac{n(n + 1)}{2} \right) = d \cdot n(n + 1) \]
Finally, the mean deviation is:
\[ D = \frac{d \cdot n(n + 1)}{2n + 1} \]

### Final Result
The mean deviation of the series \( a, a+d, a+2d, \ldots, a+2nd \) from its mean is:
\[ D = \frac{d \cdot n(n + 1)}{2n + 1} \]