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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
$\phi$
$\{99\}$
$N$
$\{9\}$
Solution
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. ' $a_1{ }^{\prime}$
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| ${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
| ${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
