Browse Questions

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

$\begin {array} {1 1} (A)\;10.0 & \quad (B)\;20.0 \\ (C)\;10.1 & \quad (D)\;20.2 \end {array}$

Toolbox:
• $S_n$ of an A.P. $=\large\frac{n}{2}$$.(2a+(n-1)d) The given series 1, 1+d, 1+2d.....1+100d is A.P. No. of terms in this series =101 Mean of this series = \overline x = \large\frac{1+(1+d)+(1+2d)+.....+(1+100d)}{101} 1+(1+d)+(1+2d)+.....+(1+100d)=\large\frac{101}{2}$$(2+(101-1)d)$
$\qquad=101(50d+1)$
$\therefore\:\overline x= \large\frac{ 101(50d+1)}{101}$$= 1+50d$
Therefore mean deviation from mean
$= \large\frac{1}{101}$ $\sum_{r=0}^{100} [ (1+rd)-(1+50d)]$
$= \large\frac{2d}{101}$ $\bigg( \large\frac{50 \times 51}{2} \bigg)$
$\Rightarrow 255 = \large\frac{50 \times 51 \times d}{101}$
$d=\large\frac{255 \times 101}{50 \times 51}$
$= 10.1$
Ans : (C)
edited Mar 26, 2014