Question

**Given:** Numbers: $k, 2k, 3k, \ldots, 1000k$ — an AP with $n = 1000$ terms; Mean Deviation about median $= 500$. **Step 1: Find the Median** Since $n = 1000$ (even), $$\text{Median} = \frac{x_{500} + x_{501}}{2} = \frac{500k + 501k}{2} = \frac{1001k}{2}$$ **Step 2: Set Up Mean Deviation Formula** $$\text{MD} = \frac{1}{1000}\sum_{i=1}^{1000}\left|ik - \frac{1001k}{2}\right|$$ **Step 3: Apply Symmetry** For $i = 1$ to $500$: $\left|ik - \dfrac{1001k}{2}\right| = \dfrac{k(1001 - 2i)}{2}$ (positive, since $2i \leq 1000 < 1001$). By symmetry of the AP about the median: $$\text{MD} = \frac{2}{1000} \cdot \sum_{i=1}^{500} \frac{k(1001 - 2i)}{2} = \frac{k}{1000}\sum_{i=1}^{500}(1001 - 2i)$$ **Step 4: Evaluate the Sum** $$\sum_{i=1}^{500}(1001 - 2i) = 1000 \cdot 1001/2 \cdot \frac{1}{1} - 2 \cdot \frac{500 \cdot 501}{2}$$ $$= 500 \times 1001 - 500 \times 501 = 500(1001 - 501) = 500 \times 500 = 250000$$ **Step 5: Solve for $k$** $$\text{MD} = \frac{k \times 250000}{1000} = 250k$$ $$250k = 500 \implies k = 2$$ **Step 6: Find $k^2$** $$k^2 = (2)^2 = 4$$ **Answer: (B)**