Question 53

Mathematics Statistics Easy

The mean deviation from the mean of the AP \( a, a+d, a+2 d, \ldots a+ \) 2nd is

(A) \( n(n+1) d \)
(B) \( \frac{n(n+1) d}{2 n+1} \)
(C) \( \frac{n(n+1) d}{2 n} \)
(D) \( \frac{n(n-1) d}{2 n+1} \)
View Dynamic Solution & Explanation
Correct Solution: Option B

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: \[ \boxed{\frac{n(n+1)d}{2n+1}} \]