Exams › JEE Advanced › Maths

Find the mean deviation about the mean of the arithmetic progression a, a + d, a + 2d,..., a + 2nd.

1. (n+1)/(2n+1) * |d|
2. n(n+1)/(2n+1) * |d|
3. n(n-1)/(2n+1) * |d|
4. none of these

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

Solution

The mean is a + nd, deviations are 0, d, 2d,..., nd on each side, summing to 2*|d|*n(n+1)/2 = |d|*n(n+1); dividing by (2n+1) gives n(n+1)/(2n+1)*|d|.

Related JEE Advanced Maths questions

• The standard deviation of a set containing 25 values is 40. If 5 is added to each value in the set, what will the updated standard deviation be?
• Given Σ(xi − 5) = 9 and Σ(xi − 5)² = 45 for a dataset of 9 values x1, x2,..., x9, what is the standard deviation of the data?
• The variables X and U are connected by the equation X = 5 + 2U. If the mean of X is 10 and its coefficient of variation is 2.6, what is the coefficient of variation for U?
• What is the average absolute deviation from the mean for the arithmetic progression a, a + d, a + 2d,..., a + 2nd?
• What is the variance of the first n positive integers?
• Seven real numbers 9 = x1 < x2 <... < x7 form an arithmetic progression with common difference d. If their standard deviation is 4 and their mean is x_bar, find the value of x_bar + x6.

⚔️ Practice JEE Advanced Maths free + battle 1v1 →