To find the mean deviation from the mean of the arithmetic sequence \( a, a+d, a+2d, \ldots, a+2nd \), we can follow these steps:
- Calculate the Mean of the Series:
The given series is an arithmetic progression with the first term \( a \) and the common difference \( d \). The number of terms is \( 2n + 1 \). The mean (\( \bar{x} \)) of an arithmetic progression is given by:
\[\bar{x} = \frac{\text{Sum of all terms}}{\text{Number of terms}} = \frac{(2n+1)a + d(1+2+3+\ldots+2n)}{2n+1}\]- Using the formula for the sum of the first \( m \) natural numbers \( 1 + 2 + \ldots + m = \frac{m(m+1)}{2} \), we find:
\[1 + 2 + \ldots + 2n = \frac{2n(2n+1)}{2} = n(2n+1)\]- Hence, the sum of the terms is:
\[(2n+1)a + dn(2n+1)\]- So the mean is:
\[\bar{x} = \frac{(2n+1)a + dn(2n+1)}{2n+1} = a + nd\]- Calculate the Mean Deviation:
Mean deviation is given by the average of absolute deviations from the mean:
\[\text{MD} = \frac{1}{2n+1} \sum_{i=0}^{2n} |(a+id) - (a+nd)|\]- This simplifies to:
\[\text{MD} = \frac{1}{2n+1} \left(\sum_{i=0}^{n-1} (nd-id) + \sum_{i=n+1}^{2n} (id-nd)\right)\]- For the range \( 0 \leq i < n \), \( |(a+id) - (a+nd)| = nd-id = d(n-i) \). Thus, we compute:
\[\sum_{i=0}^{n-1} (n-i) = n + (n-1) + \ldots + 1 = \frac{n(n+1)}{2}\]- For \( n < i \leq 2n \), \( |(a+id) - (a+nd)| = id-nd = d(i-n) \). Thus, we find:
\[\sum_{i=n+1}^{2n} (i-n) = 1 + 2 + \ldots + n = \frac{n(n+1)}{2}\]- The total deviation sum is:
\[d\left(\frac{n(n+1)}{2} + \frac{n(n+1)}{2}\right) = d \cdot n(n+1)\]- Hence, the mean deviation is:
\[\text{MD} = \frac{d \cdot n(n+1)}{2n+1}\]Therefore, the mean deviation from the mean of the series is \(\frac{n(n+1)d}{2n+1}\), which corresponds to the given correct answer.