Mean deviation of the series `a^(2), a^(2)+d, a^(2)+2d, ………………., a^(2)+2nd` from its mean is

Options

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

Answer

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

### Step 1: Identify the series and the number of terms
The series is an arithmetic progression (AP) with:
- First term \( a_1 = a^2 \)
- Common difference \( d \)
- Last term \( a^2 + 2nd \)

The number of terms \( n \) can be calculated as:
\[ \text{Number of terms} = 2n + 1 \]

### Step 2: Calculate the mean of the series
The mean \( \bar{x} \) of an AP can be calculated using the formula:
\[ \bar{x} = \frac{\text{Sum of all terms}}{\text{Number of terms}} \]

The sum of the first \( N \) terms of an AP is given by:
\[ S_N = \frac{N}{2} \times (\text{First term} + \text{Last term}) \]
Substituting the values:
\[ S_{2n+1} = \frac{2n + 1}{2} \times (a^2 + (a^2 + 2nd)) = \frac{2n + 1}{2} \times (2a^2 + 2nd) = (2n + 1)(a^2 + nd) \]

Now, substituting this into the mean formula:
\[ \bar{x} = \frac{(2n + 1)(a^2 + nd)}{2n + 1} = a^2 + nd \]

### Step 3: Calculate the mean deviation
The mean deviation \( MD \) from the mean is given by:
\[ MD = \frac{1}{N} \sum_{i=1}^{N} |x_i - \bar{x}| \]
Where \( N = 2n + 1 \) and \( x_i \) are the terms of the series.

Substituting \( \bar{x} \):
\[ MD = \frac{1}{2n + 1} \sum_{i=0}^{2n} |(a^2 + id) - (a^2 + nd)| \]
This simplifies to:
\[ MD = \frac{1}{2n + 1} \sum_{i=0}^{2n} |id - nd| = \frac{1}{2n + 1} \sum_{i=0}^{2n} |(i - n)d| \]

### Step 4: Evaluate the sum
The terms \( |(i - n)| \) will be symmetric around \( n \):
- For \( i = 0 \) to \( n \), \( |(i - n)| = n - i \)
- For \( i = n \) to \( 2n \), \( |(i - n)| = i - n \)

Thus, we can calculate the sum:
\[ \sum_{i=0}^{2n} |(i - n)| = 2 \sum_{i=0}^{n} (n - i) = 2 \left( n + (n-1) + (n-2) + \ldots + 1 + 0 \right) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

### Step 5: Substitute back into the mean deviation formula
Now substituting back:
\[ MD = \frac{1}{2n + 1} \cdot n(n + 1) = \frac{n(n + 1)}{2n + 1} \]

### Final Answer
Thus, the mean deviation of the series from its mean is:
\[ \boxed{\frac{n(n + 1)}{2n + 1}} \]