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If the maean deviation about the mean of the numbers \mathrm{1,2,3,.....,n,} where \mathrm{n} is odd,is  \mathrm{\frac{5\left ( n+1 \right )}{n}} , then \mathrm{n} is equal to ______.

Option: 1

21


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\bar{x}=\frac{1+2+3+\cdots+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\left(\frac{n+1}{2}\right)}  

                                                                       .........(Which is middle term)

\mathrm{{M.D}=\frac{\sum\left|x_{i}-\bar{x}\right|}{n}=\frac{5(n+1)}{n} }

\mathrm{\Rightarrow \frac{2\left(1+2+3+\cdots+\frac{(n-1)}{2}\right)}{n}=\frac{5(n+1)}{n}}\\

\Rightarrow \mathrm{\frac{2\left ( \frac{\left ( n-1 \right )}{2}\left ( \frac{n-1}{2}+1 \right ) \right )}{2}=\frac{5\left ( n+1 \right )}{n} }\\

\mathrm{\Rightarrow \frac{(n-1)(n+1)}{4 n}=\frac{5(n+1)}{n}}

\mathrm{n=21}

Hence the correct answer is \mathrm{n=21}

 

Posted by

Riya

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