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Let \mathrm{S} be the set of all values of \mathrm{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

Option: 1

\mathrm{N}


Option: 2

\phi


Option: 3

\{99\}


Option: 4

\{9\}


Answers (1)

best_answer

Let \mathrm{a}_{1}=\mathrm{n} \quad \mathrm{a}_{2}=\mathrm{n}+1 \quad \mathrm{a}_{3}=\mathrm{n}+2\quad \dots
\bar{x}=\frac{n+(n+1)+(n+2)+\ldots . n+99}{100}
=\frac{100 \mathrm{n}+\frac{100 \times 99}{2}}{100}=n+\frac{99}{2}

Mean deviation about the mean

\frac{1}{100} \sum\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|
\Rightarrow \frac{1}{100}\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots \ldots . .+\frac{97}{2}+\frac{99}{2}\right)
\Rightarrow \frac{2}{100}\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots \ldots .50\right.$ terms $)
\Rightarrow \frac{2}{100} \times \frac{1}{2} \times(50)^{2}=\frac{50 \times 50}{100}=25

It is 25 irrespective of the value of

\therefore \mathrm{n} \in \mathrm{N}
\Rightarrow \mathrm{S}=\mathrm{N}

Posted by

Anam Khan

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