If the mean deviation of the numbers $$1,\,1 + d,\,1 + 2d,\,......,1 + 100d$$        from their mean is $$255,$$  then $$d$$ is equal to :

Solution :
$$\eqalign{ & {\text{Mean}} = \frac{{101 + d\left( {1 + 2 + 3 + ...... + 100} \right)}}{{101}} \cr & = 1 + \frac{{d \times 100 \times 101}}{{101 \times 2}} \cr & = 1 + 50d \cr & \because \,{\text{Mean deviation from the mean}} = 255 \cr & \Rightarrow \frac{1}{{101}}\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + 2d} \right) - \left( {1 + 50d} \right)} \right| + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right] = 225 \cr & \Rightarrow 2d\left[ {1 + 2 + 3 + ...... + 50} \right] = 101 \times 225 \cr & \Rightarrow 2d \times \frac{{50 \times 51}}{2} = 101 \times 225 \cr & \Rightarrow d = \frac{{101 \times 255}}{{50 \times 51}} \cr & \Rightarrow d = 10.1 \cr} $$