JEE Main 2026 Jan 22 Shift 2, Mathematics Q2: Measure of Dispersion

JEE Main2026Jan 22, Shift 2Mathematics

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

Solution

The data is the AP $k, 2 k, \dots, 1000 k$ with median $M = \frac{1001 k}{2}$ .

Mean deviation $= \frac{1}{1000} \sum_{i = 1}^{1000} ∣ ik - M ∣$ .

By symmetry, the 500 terms on each side of the median contribute equally. Summing the deviations above the median:

$\sum = \frac{k}{2} + \frac{3 k}{2} + \frac{5 k}{2} + \dots (500 terms) = \frac{k}{2} (1 + 3 + 5 + \dots + 999) = \frac{k}{2} \cdot 50 0^{2}$ .

Doubling for both sides and dividing by 1000:

$MD = \frac{2 \cdot \frac{k}{2} ( 500 ) ^{2}}{1000} = \frac{500 k}{2} = 500$ .

So $k = 2$ and $k^{2} = 4$ .

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