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Q.
Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of $100$ consecutive positive integers $a_1, a_2, a_3, \ldots ., a_{100}$ is $25$. Then $S$ is
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{ }^{\prime} 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}$
Mean deviation about mean $=\frac{\displaystyle\sum_{ i =1}^{100}\left| x _{ i }-\overline{ x }\right|}{100}$
$=\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. ' $a _1$ '