Statistics — Medium JEE math MCQ
**[JEE Mains 2026]** Mean deviation about median for data k, 2k, 3k, ..., 1000k is 500. Find the value of k.
- A. 1
- B. 2
- C. 1/2
- D. 4
Solution
Data: k, 2k, 3k, ..., 1000k (1000 terms).
Median = average of 500th and 501st terms = (500k + 501k)/2 = 500.5k.
Mean deviation = (1/n)Σ|xᵢ - median|.
= (k/1000)[|1 - 500.5| + |2 - 500.5| + ... + |1000 - 500.5|]
= (k/1000) × 2 × [499.5 + 498.5 + ... + 0.5]
= (k/1000) × 2 × (0.5 + 1.5 + ... + 499.5)
= (k/1000) × 2 × (500 × 250) = (k/1000) × 250000 = 250k.
Given 250k = 500, so k = 2.