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

- $S_n$ 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)

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