let $a_1$ be any natural number

$a_1, a_1+1, a_1+2, \ldots ., a_1+99 \text { are values of } a_i ' S$

$\bar{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\ldots . .+a_1+99}{100}$

$=\frac{100 a_1+(1+2+\ldots . .+99)}{100}=a_1+\frac{99 \times 100}{2 \times 100}$

$=a_1+\frac{99}{2}$

$\text { Mean deviation about mean }=\frac{\sum \limits_{i=1}^{100}\left|x_i-\bar{x}\right|}{100}$

$=\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots .+\frac{1}{2}\right)}{100}$

$=\frac{1+3+\ldots .+99}{100}$

$=\frac{\frac{50}{2}[1+99]}{100}$

$=25$

So, it is true for every natural no. ' $a_1{ }^{\prime}$