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
The series is given as:
\[ a^2, a^2 + d, a^2 + 2d, \ldots, a^2 + 2nd \]

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

### Step 3: Calculate the Mean
The mean \( \bar{x} \) of the series is given by:
\[ \bar{x} = \frac{\text{Sum of all terms}}{\text{Number of terms}} = \frac{S}{2n + 1} \]
Where \( S \) is the sum of the series:
\[ S = a^2 + (a^2 + d) + (a^2 + 2d) + \ldots + (a^2 + 2nd) \]
This can be simplified as:
\[ S = (2n + 1)a^2 + d(0 + 1 + 2 + \ldots + 2n) \]
Using the formula for the sum of the first \( m \) integers:
\[ \sum_{k=0}^{m} k = \frac{m(m + 1)}{2} \]
We find:
\[ 0 + 1 + 2 + \ldots + 2n = \frac{2n(2n + 1)}{2} = n(2n + 1) \]
Thus, substituting back into \( S \):
\[ S = (2n + 1)a^2 + dn(2n + 1) \]
Now, substituting \( S \) into the mean formula:
\[ \bar{x} = \frac{(2n + 1)a^2 + dn(2n + 1)}{2n + 1} = a^2 + nd \]

### Step 4: Calculate the Mean Deviation
The mean deviation \( MD \) is defined as:
\[ MD = \frac{\sum_{i=1}^{2n+1} |x_i - \bar{x}|}{2n + 1} \]
Where \( x_i \) are the terms of the series. We can express the absolute deviations as:
\[ |x_i - \bar{x}| = |(a^2 + kd) - (a^2 + nd)| = |kd - nd| = |(k - n)d| \]
For \( k = 0, 1, 2, \ldots, 2n \), we compute:
\[ \sum_{k=0}^{2n} |(k - n)d| = d \sum_{k=0}^{2n} |k - n| \]
This can be split into two parts:
1. For \( k = 0 \) to \( n \): \( |k - n| = n - k \)
2. For \( k = n+1 \) to \( 2n \): \( |k - n| = k - n \)

Calculating each part:
- From \( k = 0 \) to \( n \):
\[ \sum_{k=0}^{n} (n - k) = n(n + 1)/2 \]
- 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 is:
\[ \sum_{k=0}^{2n} |k - n| = n(n + 1) \]
Now substituting back into the mean deviation:
\[ MD = \frac{d \cdot n(n + 1)}{2n + 1} \]

### Final Result
The mean deviation of the series from its mean is:
\[ MD = \frac{n(n + 1)d}{2n + 1} \]