If the mean deviation about the median of the numbers \(k, 2 k, 3 k, \ldots, 1000 k\) is \(500\), then \(k^2\) is equal …

Here number of terms are even therefore

median \(=\frac{500k+501k}{2}=\frac{1001\text{k}}{2}={\text{X}}_{\text{M}}\)

mean deviation about median \(=\frac{\sum \left|{\text{X}}_{\text{i}}−{\text{X}}_{\text{M}}\right|}{\text{n}}\)

\(=\frac{2\left(\frac{\text{k}}{2}+\frac{3\text{k}}{2}+\frac{5\text{k}}{2}+\ldots 500\text{ terms }\right)}{1000}\)

\(=\frac{2\times \frac{k}{2}\left(1+3+5+.....500\mathrm{terms}\right)}{1000}\\ =\frac{2\times \frac{k}{2}\left(\frac{500}{2}\left{2+\left(500-1\right)\times 2\right}\right)}{1000}\)

\(=\frac{2 \cdot \frac{k}{2}(500)^2}{1000}=\frac{500 k}{2}=500(\) given \()\)

\(\therefore k=2\)

Hence, \(k^2=4\)