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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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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

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

 

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1 Answer

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  • $S_n$ 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)
answered Jan 29, 2014 by thanvigandhi_1
edited Mar 26, 2014 by rvidyagovindarajan_1
 

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