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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Statistics

NTA Abhyas 2020 · Mathematics · Statistics

The mean deviation of the series a^ 2 , a^ 2 + d, a^ 2 + 2d,..... ,...

The mean deviation of the series a2,a2+d,a2+2d,.....,a2+2nda^{2}, a^{2} + d, a^{2} + 2d,..... , a^{2} + 2nd from its median is

Options

  1. A\dfrac{\left(n + 1\right)d}{\left(2n + 1\right)}
  2. Bnd2n+1\dfrac{nd}{2n + 1}
  3. C\dfrac{n\left(n + 1\right)d}{\left(2n + 1\right)}
  4. D\dfrac{\left(2n + 1\right)d}{n\left(n + 1\right)}

Correct answer

C. \dfrac{n\left(n + 1\right)d}{\left(2n + 1\right)}

Step-by-step solution

Median or mean (x¯)=2n+12⋅(a2+a2+2nd)(2n+1)=a2+nd\left(\overset{¯}{x}\right) = \dfrac{2n + 1}{2} ⋅ \left(a^{2} + a^{2} + 2nd\right)\left(2n + 1\right)= a^{2} + nd~ \left\{∵ Mean = Median\right\}<br/><br />Σ\left|{x}_{i} - \overset{¯}{x}\right| = 2d\left(1 + 2 + . . . . + n\right) = n\left(n + 1\right)d<br/><br />∴ M . D . = \dfrac{n\left(n + 1\right)d}{2n + 1}
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