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

Correct answer: C. 4 The sequence k, 2k, 3k, , 1000k has 1000 terms. The median is (500k + 501k)/(2) = (1001k)/(2) . The mean deviation is MD = (1)/(1000)∑_ i=1 ^ 1000 |ik - median| . For i ≤ 500 : deviation = (1001k)/(2) - ik , giving ∑_ i=1 ^ 500 ((1001)/(2) - i) = 125000 . By symmetry, the sum for i ≥ 501 is also 125000 . Thus MD = (k)/(1000) · 250000 = 250k = 500 , so k = 2 and k^2 = 4 .

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Statistics

If the mean deviation about the median of the numbers k, 2 k, 3 k,...

If the mean deviation about the median of the numbers

\mathrm{k}, 2 \mathrm{k}, 3 \mathrm{k}, \ldots. ., 1000 \mathrm{k}

is 500, then

\mathrm{k}^{2}

is equal to :

Options

Correct answer

C. 4

Step-by-step solution

The sequence

k, 2k, 3k, \ldots, 1000k

has 1000 terms.
The median is

\frac{500k + 501k}{2} = \frac{1001k}{2}

.
The mean deviation is

MD = \frac{1}{1000}\sum_{i=1}^{1000}|ik - \text{median}|

.
For

i \leq 500

: deviation =

\frac{1001k}{2} - ik

, giving

\sum_{i=1}^{500}(\frac{1001}{2} - i) = 125000

.
By symmetry, the sum for

i \geq 501

is also

125000

.
Thus

MD = \frac{k}{1000} \cdot 250000 = 250k = 500

, so

k = 2

and

k^2 = 4

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