Question

The mean deviation and S.D about actual mean of the series $a,a+d,a+2d,.....,a+\left(2n\right)d$ are respectively

• A. $\frac{n(n+1)d}{2n+1},\sqrt{\frac{n(n-1)}{3}}d$
• B. $\frac{n(n-1)}{3},\frac{n(n+1)}{2n}d$
• C. $\frac{n(n+1)d}{2n+1},\sqrt{\frac{n(n+1)}{3}}d$
• D. $\frac{n(n-1)d}{2n-1},\sqrt{\frac{n(n-1)}{3}}d$

Answer 1

Answer: (C)

$\frac{n(n+1)d}{2n+1},\sqrt{\frac{n(n+1)}{3}}d$

$\overline{x}=$ Mean of the series $=\frac{a+(a+d)+.....+(a+2nd)}{2n+1}=a+nd$
$\begin{array}{ccc}{x}_i& d=|x_i-\overline{x}|& {\left|d\right|}^2=D\\ a& nd& n^2{d}^2\\ a+d& (n-1)d& {(n-1)}^2{d}^2\end{array}$
$\begin{array}{ccc}a+(n-2)& 2d& 4d^2\\ a+(n-1)d& 1d& d^2\\ a+nd& 0& 0\\ a+(n+1)d& 1d& d^2\\ a+(n+2)d& 2d& 4d^2\end{array}$
$a+2nd\frac{nd}{\sum \left|\alpha \right|=2nd}\left(\frac{n+1}{2}\right),\frac{n^2{d}^2}{\sum {\left|\alpha \right|}^2}=2d^2$
solutions is not unbersting $[1^2+2^2+...+n^2]$
$=n(n+1)d=\frac{2d^2\left(n\right)(n+1)(2n+1)}{6}$
now $M.D=\frac{\sum \left|d\right|}{n=\sum f}and{\sigma }^2=\frac{\sum {\left|d\right|}^2}{\sum f=N}$
$M.D=\frac{n(n+1)d}{2n+d}$ and ${\sigma }^2=\frac{2d^2.n(n+1)(2n+1)}{6.(2n+1)}$
$=\frac{n(n+1)d^2}{3}$
$\therefore S.D=\sqrt{{\sigma }^2}=\sqrt{\frac{n(n+1)}{3}}.d$