Find the mean deviation from the mean and standard deviation of the series:

a, a+d, a+2d, ....., (a+nd), .... a+(2n-1)d, a+2nd

its urgent..

Question

Find the mean deviation from the mean and standard deviation of
the series:
a, a+d, a+2d, , (a+nd), a+(2n-1)d, a+2nd
its urgent

Ankush Jain · Accepted Answer

Given series is: a, a+d, a+2d, , (a+nd), + a+(2n-1)d + (a + 2nd) Here, the last term in the series is (a + 2nd) Clearly, it is (2n + 1) term in the given series Therefore, $M e a n = \bar{x} = \frac{1}{2 n + 1} (a + a + d + a + 2 d + + a + 2 n d) = \frac{1}{2 n + 1} \times (\frac{2 n + 1}{2}) [2 a + (2 n + 1 - 1) d] = \frac{1}{2} [2 a + 2 n d] = a + n d$ Now, mean deviation about the mean $= \frac{1}{2 n + 1} \sum_{p = 1}^{2 n} [(a + p d) - (a + n d)] = \frac{1}{2 n + 1} [2 n d + 2 (n - 1) d + 2 d] = \frac{2}{2 n + 1} [n d + (n - 1) d + + d] = \frac{n (n + 1) d}{2 n + 1}$ Now, $σ^{2} = \frac{1}{2 n + 1} \sum_{p = 0}^{2 n} {(a + p d) - (a + n d)}^{2} = \frac{1}{2 n + 1} {2 n^{2} d^{2} + 2 d^{2} (n - 1)^{2} + 2 d^{2} (n - 2)^{2} + + {2 d}^{2}} = \frac{2 d^{2}}{2 n + 1} {n^{2} + (n - 1)^{2} + (n - 2)^{2} + + 1^{2}} = \frac{2 d^{2}}{2 n + 1} \frac{n (n + 1) (2 n + 1)}{6} = \frac{n (n + 1) d^{2}}{3} ∴ s \tan d a r d d e v i a t i o n = σ = d \sqrt{\frac{n (n + 1)}{3}}$

Find the mean deviation from the mean and standard deviation of the series: a, a+d, a+2d, ....., (a+nd), .... a+(2n-1)d, a+2nd its urgent..

Find the mean deviation from the mean and standard deviation of the series:

a, a+d, a+2d, ....., (a+nd), .... a+(2n-1)d, a+2nd

its urgent..