Question

he mean deviation of the series a, a+d, a+2d, ……, a+2n from its mean is

A. \(\frac{(n+1)d}{2n+1}\)

B. \(\frac{nd}{2n+1}\) 

C. \(\frac{n(n+1)d}{2n+1}\) 

D. \(\frac{(2n+1)d}{n(n+1)}\) 

Answer 1

Given, Series is a, a+d, a+2d,…,+a+2n

Mean (X) = \(\frac{a+a+d+a+2d+....+a+2nd}{2n+1}\) 

\(\bar X\) = a + nd

Now, deviation from mean is x_i - X

Hence, M.D. =  \(\frac{n(n+1)d}{2n+1}\)