The mean deviation from the mean of the AP \( a, a+d, a+2 d, \ldots a+ \) 2nd is
Step-by-step Solution:
Solution: Mean Deviation from the Mean of an A.P.
Given Arithmetic Progression
The given arithmetic progression (A.P.) is:
\[
a, a + d, a + 2d, \dots, a + 2nd
\]
It has \( 2n + 1 \) terms.
Mean of the A.P.
The formula for the mean of an arithmetic progression is given by:
\[
\bar{X} = \frac{\text{Sum of all terms}}{\text{Number of terms}}
\]
The sum of the terms in an A.P. is:
\[
S = \sum_{k=0}^{2n} (a + kd)
\]
Using the summation formula:
\[
S = (2n + 1) a + d \sum_{k=0}^{2n} k
\]
Since,
\[
\sum_{k=0}^{2n} k = \frac{(2n)(2n+1)}{2} = n(2n+1)
\]
we get,
\[
S = (2n+1) a + d n(2n+1)
\]
Thus, the mean is:
\[
\bar{X} = \frac{S}{2n+1} = \frac{(2n+1)a + n(2n+1)d}{2n+1}
\]
\[
\bar{X} = a + nd
\]
Mean Deviation Formula
The mean deviation from the mean is given by:
\[
M.D = \frac{1}{2n+1} \sum_{k=0}^{2n} |(a + kd) - (a + nd)|
\]
\[
= \frac{1}{2n+1} \sum_{k=0}^{2n} |kd - nd|
\]
\[
= \frac{1}{2n+1} \sum_{k=0}^{2n} |(k - n)d|
\]
Since \( |k - n| \) is symmetric about \( n \), the sum simplifies to:
\[
\sum_{k=0}^{2n} |k - n| = n(n+1)
\]
Thus, the mean deviation is:
\[
M.D = \frac{n(n+1)d}{2n+1}
\]
Correct Answer
The correct answer is:
\[
\boxed{\frac{n(n+1)d}{2n+1}}
\]