Question 13 Marks
Find the mean deviation and standard deviation of the series $a, a+d, a+2 d, \ldots, a+2 n d$ from mean and prove that the variable is greater than the first one.
Answer
View full question & answer→Given series $a, a+d, a+2 d, \ldots a+2 n d$ is an arithmetic progression.
$\begin{aligned}\text {So, its mean}\quad\bar{x} & =\frac{\text { First term }+ \text { last term }}{2} \\& =\frac{a+a+2 n d}{2}=\frac{2(a+n d)}{2} \\\bar{x} & =(a+n d)\end{aligned}$
We know that :
$\begin{array}{l}\text { Mean deviation M.D. }=\frac{\sum|x-\bar{x}|}{n} \\\text { Here number of terms }=(2 n+1)\end{array}$
$\therefore \quad\quad\quad\quad\quad\quad\ \text { M.D. }=\frac{\sum\left|x_i-\bar{x}\right|}{2 n+1}$
$\begin{array}{r}\{a-(a+n d)\}+\{a+d-(a+n d)\}+ \{a+2 d-(a+n d)\}+\ldots \ldots\end{array}$
$\text { M.D. }=\frac{\{(a+2 n d)-(a+n d)\}}{2 n+1}$
$=\frac{\begin{array}{r}(-n d)+d(1-n)+d(2-n) \ldots \ldots \\d(n-n)+d(n+1-n)+ \\d(n+2-n)+\ldots \ldots d(n+n-n)\end{array}}{2 n+1}$
$=\frac{2 d(1+2+3+4+\ldots \ldots n)}{(2 n+1)}=\frac{2 d \frac{n(n+1)}{2}}{(2 n+1)}$
Since sum of $n$ terms of natural numbers $=\frac{n(n+1)}{2}$
$\Rightarrow \quad \text { M.D. }=\frac{n(n+1) d}{(2 n+1)}\ldots\ldots (1)$
Standard deviation
$\sigma=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}$
Here, $n=(2 n+1)$
$\therefore \sigma=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{(2 n+1)}}$
$=\sqrt{\begin{array}{l}(-n d)^2+\{(n-1) d\}^2+\{(n-2) d\}^2+\ldots \ldots . .+ \\ \frac{\{(n-n) d\}^2+\{n-(n+1) d\}^2+\ldots \ldots\left\{\{n-(n+n) d\}^2 \right.}{(2 n+1)}\end{array}}$
The above expression can be written as :
$\sigma=\sqrt{\frac{2 d^2\left\{1^2+2^2+3^2+4^2+\ldots \ldots+n^2\right\}}{(2 n+1)}}$
As we know that sum of arithmetic progression
$\left(1^2+2^2+3^2+4^2+\ldots+n^2\right) \text { is } \frac{n(n+1)(2 n+1)}{6}$
Putting values,
$\begin{array}{l}\sigma=\sqrt{\frac{2 d^2(n)(n+1)(2 n+1)}{6 \times(2 n+1)}} \\ \sigma=d \sqrt{\frac{n(n+1)}{3}}\ldots\ldots (2)\end{array}$
It is clear from equation (1) and (2), the variable is greater than the first one.
Hence proved.
$\begin{aligned}\text {So, its mean}\quad\bar{x} & =\frac{\text { First term }+ \text { last term }}{2} \\& =\frac{a+a+2 n d}{2}=\frac{2(a+n d)}{2} \\\bar{x} & =(a+n d)\end{aligned}$
We know that :
$\begin{array}{l}\text { Mean deviation M.D. }=\frac{\sum|x-\bar{x}|}{n} \\\text { Here number of terms }=(2 n+1)\end{array}$
$\therefore \quad\quad\quad\quad\quad\quad\ \text { M.D. }=\frac{\sum\left|x_i-\bar{x}\right|}{2 n+1}$
$\begin{array}{r}\{a-(a+n d)\}+\{a+d-(a+n d)\}+ \{a+2 d-(a+n d)\}+\ldots \ldots\end{array}$
$\text { M.D. }=\frac{\{(a+2 n d)-(a+n d)\}}{2 n+1}$
$=\frac{\begin{array}{r}(-n d)+d(1-n)+d(2-n) \ldots \ldots \\d(n-n)+d(n+1-n)+ \\d(n+2-n)+\ldots \ldots d(n+n-n)\end{array}}{2 n+1}$
$=\frac{2 d(1+2+3+4+\ldots \ldots n)}{(2 n+1)}=\frac{2 d \frac{n(n+1)}{2}}{(2 n+1)}$
Since sum of $n$ terms of natural numbers $=\frac{n(n+1)}{2}$
$\Rightarrow \quad \text { M.D. }=\frac{n(n+1) d}{(2 n+1)}\ldots\ldots (1)$
Standard deviation
$\sigma=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{n}}$
Here, $n=(2 n+1)$
$\therefore \sigma=\sqrt{\frac{\sum\left(x_i-\bar{x}\right)^2}{(2 n+1)}}$
$=\sqrt{\begin{array}{l}(-n d)^2+\{(n-1) d\}^2+\{(n-2) d\}^2+\ldots \ldots . .+ \\ \frac{\{(n-n) d\}^2+\{n-(n+1) d\}^2+\ldots \ldots\left\{\{n-(n+n) d\}^2 \right.}{(2 n+1)}\end{array}}$
The above expression can be written as :
$\sigma=\sqrt{\frac{2 d^2\left\{1^2+2^2+3^2+4^2+\ldots \ldots+n^2\right\}}{(2 n+1)}}$
As we know that sum of arithmetic progression
$\left(1^2+2^2+3^2+4^2+\ldots+n^2\right) \text { is } \frac{n(n+1)(2 n+1)}{6}$
Putting values,
$\begin{array}{l}\sigma=\sqrt{\frac{2 d^2(n)(n+1)(2 n+1)}{6 \times(2 n+1)}} \\ \sigma=d \sqrt{\frac{n(n+1)}{3}}\ldots\ldots (2)\end{array}$
It is clear from equation (1) and (2), the variable is greater than the first one.
Hence proved.