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  • The mean deviation from the mean of the AP a, a + d, a + 2d, …., a + 2nd is:Option: 1 <img alt="\mathrm{\mathrm{n(n+1)d}}" src="http

The mean deviation from the mean of the AP a, a + d, a + 2d, …., a + 2nd is:

Option: 1

\mathrm{\mathrm{n(n+1)d}}

 

 

 

 


Option: 2

\mathrm{ \frac{n(n+1)d}{2n+1}}
 


Option: 3

\mathrm{\frac{n(n+1)d}{2n}}

 


Option: 4

\mathrm{ \frac{n(n-1)d}{2n+1}}


Answers (1)

best_answer

The correct answer is b) \mathrm{ \frac{n(n+1)d}{2n+1}}

The mean of the series a, a + d, a + 2d, …., a + 2nd is:

\mathrm{\overline{X} = \frac{1}{2n+1}\left [ a + a+d+a+2d+...+a+2nd \right ]}

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

Therefore, the Mean deviation from the mean 

\begin{aligned} & =\frac{1}{2 n+1} \sum_0^{2 n}|(a+r d)-(a+n d)| \\ & =\frac{1}{2 n+1} \sum_0^{2 n}|r-n| d \\ & =\frac{1}{2 n+1} 2 d(1+2+\ldots+n) \\ & =\frac{n(n+1) d}{2 n+1} \end{aligned}

Therefore b) is the correct answer.

 

     

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