1.The mean deviation of the numbers a,a+d,a+2d,…,a+2nda, a+d, a+2d, \ldots, a+2nda,a+d,a+2d,…,a+2nd from their mean is equal toa.(n+1)d2n+1\frac{(n+1)d}{2n+1}2n+1(n+1)db.n(n+1)d2n+1\frac{n(n+1)d}{2n+1}2n+1n(n+1)dc.(n+1)∣d∣2n\frac{(n+1)|d|}{2n}2n(n+1)∣d∣d.n(n+1)∣d∣2n+1\frac{n(n+1)|d|}{2n+1}2n+1n(n+1)∣d∣Login to continueOnly logged in users canattempt or see the solution.
