The mean deviation of items $ \displaystyle x, x + y, x + 2y, ... x + 2ny $ from mean is
- $ \displaystyle \frac{n\left ( n+1\right )y}{2n+1} $
- $ \displaystyle \frac{\left ( n+1\right )y}{2n+1} $
- $ \displaystyle\frac{ny}{2n+1} $
- $ \displaystyle\frac{\left ( 2n+1 \right )y} {n\left ( n+1 \right )} $
The series is an arithmetic progression with 2n+1 terms: x, x+y, ..., x+2ny. The mean is x+ny. The mean deviation is the average of the absolute differences from the mean: (1/(2n+1)) * sum(|(x+ky) - (x+ny)|) for k=0 to 2n. This simplifies to (y/(2n+1)) * (sum(|k-n|)). The sum is 2 * (1+2+...+n) = 2 * (n(n+1)/2) = n(n+1). Thus, the mean deviation is n(n+1)y / (2n+1).
The given observations form an arithmetic progression of 2n plus 1 terms with a mean of x plus ny. The absolute deviations of the terms from the mean are the multiples of y from 0 up to 2ny, so the sum of these absolute deviations is y multiplied by twice the sum of the first n natural numbers, giving 2y times n(n+1)/2, which simplifies to n(n+1)y. Dividing this total deviation by the number of terms (2n+1) gives the mean deviation as n(n+1)y divided by (2n+1).