If the mean deviation about the median of the numbers k,2k,3k, .,1000k is 500 , then k^2 i…

Question

If the mean deviation about the median of the numbers $k,2k,3k,\ldots .,1000k$ is 500 , then ${k}^{2}$ is equal to :

Accepted Answer

The median of $k, 2k, \ldots, 1000k$ (1000 terms) is $\dfrac{1001k}{2}$ ... actually the median (average of 500th and 501st terms) is $X_M = \dfrac{1000k+1001k}{2}\cdot\frac12$, which simplifies (as given) to $X_M = \dfrac{1001k}{2}$.

Mean deviation about the median:
$$\text{M.D.} = \frac{\sum |X_i - X_M|}{n} = \frac{2\left(\frac{k}{2}+\frac{3k}{2}+\frac{5k}{2}+\cdots \text{(500 terms)}\right)}{1000} = \frac{2\cdot\frac{k}{2}(500)^2}{1000} = \frac{500k}{2}$$

Given this equals 500: $\dfrac{500k}{2} = 500 \Rightarrow k = 2 \Rightarrow k^2 = 4$.

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