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