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For the arithmetic progression a, a + d, a + 2d,..., a + 2nd, what is the mean deviation about the mean?

1. n(n+1)d
2. n(n+1)d/(2n+1)
3. n(n+1)d/(2n)
4. n(n-1)d/(2n+1)

Correct answer: n(n+1)d/(2n+1)

Solution

The AP a, a+d, ..., a+2nd has (2n+1) terms with mean a+nd. The deviations are |k-n|*d for k=0..2n, summing to 2*d*(1+2+...+n) = n(n+1)d. Dividing by (2n+1) terms gives mean deviation = n(n+1)d/(2n+1).

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