NIMCET 2014 Question Paper with solutions

Given, AP is
a,a + d, a + 2d, ...,a + 2nd
Mean of the AP
$\stackrel{-}{\mathit{x}}$ = $\frac{1}{2\mathit{n}+1}$ [a + a + d + ... + a + 2nd]
= $\frac{1}{2\mathit{n}+1}$ $\left[\frac{2\mathit{n}+1}{2}\left(\mathit{a}+\mathit{a}+2\mathit{n}\mathit{d}\right)\right]$
= a + nd
∴ Mean deviation from the mean = $\frac{1}{2\mathit{n}+1}$ $\underset{\mathit{r}=0}{\overset{2\mathit{\pi }}{\mathit{\Sigma }}}$ |(a + rd) - (a + rd)|
= $\frac{1}{2\mathit{n}+1}$ $\underset{\mathit{r}=0}{\overset{2\mathit{n}}{\mathit{\Sigma }}}$ |(r - 1n d|
= $\frac{1}{2\mathit{n}+1}$ 2d (1 + 2 + 3 + ... + n)
= $\frac{1}{2\mathit{n}+1}$ $\frac{2\mathit{d}\left(\mathit{n}\left(\mathit{n}+1\right)\right)}{2}$
= $\frac{\mathit{n}\left(\mathit{n}+1\right)\mathit{d}}{2\mathit{n}+1}$