The mean deviation of the series `a^(2), a^(2)+d, a^(2)+2d, ….., a^(2)+2nd` from its median 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 median, we will follow these steps:

### Step 1: Identify the series and its terms
The series is given as:
\[ a^2, a^2 + d, a^2 + 2d, \ldots, a^2 + 2nd \]
This is an arithmetic progression (AP) with the first term \( a^2 \) and common difference \( d \).

### Step 2: Determine the number of terms in the series
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 \( k = 2n \)).
- Therefore, the total number of terms is \( 2n + 1 \).

### Step 3: Calculate the median of the series
For an odd number of terms (which is \( 2n + 1 \)), the median is the middle term. The middle term can be found as:
\[ \text{Median} = \text{Term at position } (n + 1) \]
This term is:
\[ a^2 + nd \]

### Step 4: Calculate the mean deviation from the median
The mean deviation from the median is given by:
\[ \text{Mean Deviation} = \frac{1}{N} \sum |x_i - \text{Median}| \]
where \( N \) is the total number of terms.

### Step 5: Calculate the absolute deviations
Now we need to find the absolute deviations from the median \( a^2 + nd \):
1. For the first term:
\[ |a^2 - (a^2 + nd)| = | -nd | = nd \]
2. For the second term:
\[ |(a^2 + d) - (a^2 + nd)| = |d - nd| = |(1-n)d| \]
3. Continuing this way, we can express the absolute deviations for all terms.

The deviations will be:
- \( nd \)
- \( |(1-n)d| \)
- \( |(2-n)d| \)
- ...
- \( |(2n-n)d| = |(n)d| \)

### Step 6: Sum the absolute deviations
The total sum of these deviations can be calculated:
\[ S = nd + |(1-n)d| + |(2-n)d| + \ldots + |(2n-n)d| \]

### Step 7: Calculate the mean deviation
The mean deviation is then:
\[ \text{Mean Deviation} = \frac{S}{2n + 1} \]

### Final Result
After calculating the sum \( S \) and dividing by \( 2n + 1 \), we find that the mean deviation of the series from its median simplifies to:
\[ \frac{n(n+1)d}{2n + 1} \]