Question

The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is

पर्याय

•  \[\frac{(n + 1) d}{2n + 1}\]

   

• \[\frac{nd}{2n + 1}\]

   

•  \[\frac{n (n + 1) d}{2n + 1}\]

   

• \[\frac{(2n + 1) d}{n (n + 1)}\]

   

Answer 1

\[\frac{n (n + 1) d}{2n + 1}\]

\[x_i\]	\[\left| x_i - X \right| = \left| x_i - \left( a + nd \right) \right|\]
a	nd
a + d	(n-1)d
a + 2d	(n-2)d
a + 3d	(n-3)d
:	:
:	:
a + (n - 1)d	d
a + nd	0
a + (n+1)d	d
:	:
:	:
a + 2nd	nd
\[\sum_{} x_i = \left( 2n + 1 \right)\left( a + nd \right)\]	\[\sum_{} \left| x_i - X \right| = n\left( n + 1 \right)d\]

\[\text{ There are 2n + 1 terms } . \]

\[ \Rightarrow N = 2n + 1\]

\[ \sum_{} x_i = a + a + d + a + 2d + a + 3d + . . . + a + 2nd\]

\[ = (2n + 1)a + d (1 + 2 + 3 + . . . + 2n) \left[ a + a + a + . . . (2n + 1) \text{ times }  = \left( 2n + 1 \right) a \right]\]

\[ = (2n + 1)a + \frac{2n\left( 2n + 1 \right)d}{2} \left[ \text{ Sum of the first n natural numbers is }  \frac{n (n + 1)}{2}, \text{ but here we are considering the first 2n numbers } . \right]\]

\[ = \left( 2n + 1 \right)a + \left( 2n + 1 \right)nd \]

\[ = \left( 2n + 1 \right)\left( a + nd \right)\]

\[X = \frac{\left( 2n + 1 \right)\left( a + nd \right)}{\left( 2n + 1 \right)}\]

\[ = a + nd\]

\[ \sum_{} \left| x_i - X \right| = nd + (n - 1)d + (n - 2)d + . . . + d + 0 + d + 2 d + 3d + . . . + nd\]

\[ = d\left( n + \left( n - 1 \right) + \left( n - 2 \right) + . . . + 1 \right) + 0 + d\left( 1 + 2 + 3 + . . . + n \right)\]

\[ = \frac{dn\left( n + 1 \right)}{2} + \frac{dn\left( n + 1 \right)}{2} \left\{ \because 1 + 2 + 3 + . . . + n = \frac{n\left( n + 1 \right)}{2} \right\}\]

\[ = n(n + 1)d\]

\[\text{ Mean deviation about the mean } = \frac{\sum_{} \left| x_i - X \right|}{N}\]

\[ = \frac{n(n + 1)d}{\left( 2n + 1 \right)}\]