If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ....., 1 + 100d from their mean is 255, then the value of d is
There are 101 terms: \(1 + kd\) for \(k = 0\) to \(100\). The mean is \(1 + 50d\).
Mean deviation = \(\frac{1}{101}\sum_{k=0}^{100}|k - 50|\,d = \frac{d}{101} \times 2(1 + 2 + \dots + 50) = \frac{2550d}{101}\)
\(\frac{2550d}{101} = 255 \Rightarrow d = \frac{255 \times 101}{2550} = 10.1\)
Answer: (B) 10.1
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