Question

If the mean deviation about the mean of the numbers 1, 2, 3, …. n, where n is odd, is \(\frac{5(n + 1)}{n}\), then n is equal to ______.

Answer 1

Correct Answer: 21

The mean deviation about the mean for a sequence of consecutive numbers \(1, 2, 3, \ldots, n\) can be calculated. First, determine the mean \(\bar{x}\) of the sequence. For an arithmetic sequence with first term \(a\) and last term \(l\), the mean is given by:

\(\bar{x} = \frac{a + l}{2} = \frac{1 + n}{2}\)

The mean deviation about the mean is defined by:

\(\frac{\sum |x_i - \bar{x}|}{n}\)

Given that the mean deviation is \(\frac{5(n + 1)}{n}\), equate and solve the equation:

\(\frac{\sum |x_i - \frac{n + 1}{2}|}{n} = \frac{5(n + 1)}{n}\)

When calculating \(\sum |x_i - \frac{n + 1}{2}|\), symmetry allows simplification. Since \(n\) is odd, the sequence is symmetric around the mean. This results in pairs like \((1, n), (2, n-1), \ldots\) with each contributing the same distance from the mean.

For the mean deviation:

\(\sum |x_i - \frac{n + 1}{2}| = \frac{n^2 - 1}{4}\)

Thus, we have:

\(\frac{\frac{n^2 - 1}{4}}{n} = \frac{5(n + 1)}{n}\)

Simplify and solve for \(n\):

\(\frac{n^2 - 1}{4n} = \frac{5(n + 1)}{n}\)

Clear the fraction by multiplying through by \(4n\):

\(n^2 - 1 = 20(n + 1)\)

Expanding and rearranging gives:

\(n^2 - 20n - 21 = 0\)

Use the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -20\), \(c = -21\):

\(n = \frac{20 \pm \sqrt{400 + 84}}{2}\)

\(n = \frac{20 \pm \sqrt{484}}{2}\)

\(n = \frac{20 \pm 22}{2}\)

\(n = 21\) or \(-1\)

The possible value \(n = 21\) is valid since \(n\) must be a positive integer and lies within the range [21,21]. Thus, \(n\) is indeed 21.