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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 -

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 -

Grade:11

2 Answers

Arun
25763 Points
one year ago
Sₙ of an AP = n/2[2a + (n - 1)d]
The given series 1 , 1 + d, 1 + 2d, ........ 1 + 100d
Number of terms in the series = 101
Mean of this series = S₁₀₁ of [1  + 1 + d + 1 + 2d .....+ (1 + 100d)]/101.
Sum to 101 terms of series 1  + 1 + d + 1 + 2d .....+ (1 + 100d)
                                                     = S₁₀₁ =  101/2[2*1 + (101 - 1)d]
                                                    = 101 (1 + 50d)
=> Mean of the series = 101 (1 + 50d)/101 = 1 + 50d.
Therefore mean deviation from mean
=  1/101  *   101
                 ∑    (1 + rd) - (1 + 50d)
                r = 0
             =  2d/101 ( 50 * 51 /2)
=> 255 = 2d/101 ( 50*51/2)
=> 255 = 50 * 51 * d / 101
=> d = 255 * 101 / 50 * 51 = 10.1
 
Vikas TU
14149 Points
one year ago
| (x - mean)/n | =  255
Mean = (1 + 1 + 2d + 1 + 3d + ... +1+ 100d)/101
= 101 + d(100x101/2)/101
= 1 + (50x101d)/101
= 1+ 50d
|x - mean| = 255 x 101
The correct option is B.

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