Let S be the set of all values of a _ 1 for which the mean...

Given,

${a}_{1}$ be any natural number

So, ${a}_{1}, {a}_{1} + 1, {a}_{1} + 2, \ldots . , {a}_{1} + 99$ are values of ${a}_{i}$ or observation.

Now mean of the observation is given by,

$\overset{¯}{x} = \dfrac{{a}_{1} + \left({a}_{1} + 1\right) + \left({a}_{1} + 2\right) + \ldots . + {a}_{1} + 99}{100}$

$⇒ \overset{¯}{x} = \dfrac{100{a}_{1} + ( 1 + 2 + \ldots . . + 99 )}{100}$

$⇒ \overset{¯}{x} = {a}_{1} + \dfrac{99 \times 100}{2 \times 100}$

$⇒ \overset{¯}{x} = {a}_{1} + \dfrac{99}{2}$

Mean deviation about mean$= \dfrac{{∑}_{i = 1}^{100}\left|{x}_{i} - \overset{¯}{x}\right|}{100}$

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

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

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

$= 25$

So, it is true for every natural no. ' ${a}_{1}$ ' as there is no term of ${a}_{1}$ in mean deviation.