AP EAMCET · Maths · Statistics
The mean deviation of the numbers \(a, a+d, a+2 d, \ldots, a+2 n d\) from their mean is equal to
- A \(\frac{(n+1) d}{2 n+1}\)
- B \(\frac{n(n+1) d}{2 n+1}\)
- C \(\frac{(n+1)|d|}{2 n}\)
- D \(\frac{n(n+1)|d|}{2 n+1}\)
Answer & Solution
Correct Answer
(D) \(\frac{n(n+1)|d|}{2 n+1}\)
Step-by-step Solution
Detailed explanation
\(\bar{x} = \frac{a + (a+2nd)}{2} = a+nd\) \(\sum |x_i - \bar{x}| = 2 \sum_{k=1}^{n} |a+kd - (a+nd)| = 2 \sum_{k=1}^{n} |(k-n)d|\) \(= 2|d| \sum_{k=1}^{n} |k-n| = 2|d| \sum_{j=1}^{n} j = 2|d| \frac{n(n+1)}{2} = n(n+1)|d|\) \(N = 2n+1\)…
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