Question

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

• A. $\{99\}$
• B. $\phi$
• C. $N$
• D. $\{9\}$

Answer 1

The Correct Option is C

The correct answer is (C) : $N$
let $a_1$ be any natural number
$a_1,a_1+1,a_1+2,\dots .,a_1+99\text{are values of}{a}_i{}^{\prime }S$
$\bar{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\dots ..+a_1+99}{100}$
$=\frac{100a_1+(1+2+\dots ..+99)}{100}=a_1+\frac{99\times 100}{2\times 100}$
$=a_1+\frac{99}{2}$
Mean deviation about mean $=\frac{\sum _{i=1}^{100}\left|x_i-\overline{x}\right|}{100}$
$=\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\dots .+\frac{1}{2}\right)}{100}$
$=\frac{1+3+\dots .+99}{100}$
$=\frac{\frac{50}{2}[1+99]}{100}$
$=25$
So, it is true for every natural no. ' $a_1$ '