Mean deviation of the series `a^(2), a^(2)+d, a^(2)+2d, ………………., a^(2)+2nd` from its mean is

The mean deviation of the series a^(2), a^(2)+d, a^(2)+2d, ….., a^(2)+2nd from its median is

The mean deviation of the series a , a+d , a+2d ,.....,a+2nd from its mean is (a) ((n+1)d)/(2n+1) (b) (n d)/(2n+1) (c) (n(n+1)d)/(2n+1) (d) ((2n+1)d)/(n(n+1))

The mean deviation of the series a , a+d , a+2d , , a+2n from its mean is ((n+1)d)/(2n+1) (b) (n d)/(2n+1) (c) (n(n+1)d)/(2n+1) (d) ((2n+1)d)/(n(n+1))

Mean deviation of the series a,a+d,a+2d,......,a+2nd from its mean is (a) ((n+1)d)/((2n+1))( b )(nd)/(2n+1) (c) ((2n+1)d)/(n(n+1)) (d) (n(n+1)d)/(2n+1)

The mean deviation of the series a , a+d , a+2d ,.....,a+2n from its mean is (a) ((n+1)d)/(2n+1) (b) (n d)/(2n+1) (c) (n(n+1)d)/(2n+1) (d) ((2n+1)d)/(n(n+1))

The mean deviation of the series a, a+d, a+2d, . . . , a+2n from its mean is -

mean of the series a , a+d, a+2d,……………, a + 2nd is …..

The mean of the series a,a+d, a+2d,…,a+2nd is

The mean of the series a, a + d, a + 2d, …, a + 2 nd, is

If the mean deviation of the numbers : 1, 1 + d, 1 + 2d, ….., 1 + 100d from their mean is 255, then d is equal to :