Q55 / 170 · Manipal (MET) 2015 · Mathematics
The mean deviation from the mean of the series $a, a+d, a+2 d, \ldots, a+2 n d$, is
- A$n(n+1) d$
- Correct answerB$\frac{n(n+1) d}{2 n+1}$
- C$\frac{n_i(n+1) d}{2 n}$
- D$\frac{n(n-1) d}{2 n+1}$
Solution
The mean of the series $a, a+d, a+2 d, \ldots a+2 n d$
$\begin{aligned}
& \text { is } \bar{x}=\frac{1}{2 n+1}[a+a+d+a+2 d+\ldots+a+2 n d] \\
& \Rightarrow \bar{x}=\frac{1}{2 n+1}\left\{\frac{2 n+1}{2}(a+a+2 n d)\right\}=a+n d
\end{aligned}$
$\therefore$ Mean deviation from mean,
$\begin{aligned} & \mathrm{MD}=\frac{1}{2 n+1} \sum_{r=0}^{2 n}|(a+r d)-(a+n d)| \\ \Rightarrow \quad & \mathrm{MD}=\frac{1}{2 n+1} \sum_{r=0}^{2 n}|r-n| d \\ \Rightarrow \quad & \mathrm{MD}=\frac{1}{2 n+1}\{2 d(1+2+\ldots+n)\}=\frac{n(n+1) d}{2 n+1}\end{aligned}$
Hence, the correct answer is option: 'B'.