Question

The mean deviation from the mean of the series \(a, a+d, a+2d, .........., a+2nd\), is

• A. \(n(n+1)d\)
• B. \(\frac{n(n+1)d}{2n+1}\)
• C. \(\frac{n(n+1)d}{2n}\)
• D. \(\frac{n(n-1)d}{2n+1}\)

Hint:

For symmetric data about the mean, sum of absolute deviations is twice the sum of positive deviations.

Answer 1

The Correct Option is B

To find the mean deviation from the mean of the given arithmetic series \(a, a+d, a+2d, \ldots, a+2nd\), we will follow these steps:

1. First, determine the mean of the series. The series is an arithmetic progression with the first term \(a\), common difference \(d\), and number of terms \(2n+1\).
2. The mean \(M\) of an arithmetic progression (AP) is given by the formula: M = \frac{\text{Sum of all terms}}{\text{Number of terms}}
3. The middle term of the series (since it has \(2n+1\) terms) can be found as: T_{middle} = a + nd So, the mean \(M\) is simply this middle term, M = a + nd
4. Next, calculate the deviation of each term from the mean, and then determine the absolute sum of these deviations to find the mean deviation.
5. The mean deviation \(MD\) from the mean in an arithmetic progression is given by: MD = \frac{2}{2n+1} \left( d (1 + 2 + 3 + \ldots + n) \right)
6. The sum of the first \(n\) natural numbers is: 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}
7. Substituting this in the formula for \(MD\), we get: MD = \frac{2d}{2n+1} \times \frac{n(n+1)}{2} Simplifying, MD = \frac{n(n+1)d}{2n+1}

Therefore, the mean deviation from the mean of the series is \frac{n(n+1)d}{2n+1}. This matches the correct answer given in the options.