Statistics – Mean Deviation – JEE Main 22 Jan 2026 Shift 2

Solution:

The given numbers are $k, 2k, 3k, …, 1000k$.
Here, the number of terms $n = 1000$ (even).

The median ($M$) of an AP with an even number of terms is the average of the two middle terms.
$$ M = \frac{\left(\frac{n}{2}\right)^{th} \text{term} + \left(\frac{n}{2} + 1\right)^{th} \text{term}}{2} $$

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

The formula for Mean Deviation about Median is:
$$ M.D.(M) = \frac{\sum |x_i – M|}{n} $$

Since the terms are symmetric about the median, we can sum the deviations of the first 500 terms and multiply by 2.
$$ \sum |x_i – M| = 2 \times \left( \left| k – \frac{1001k}{2} \right| + \left| 2k – \frac{1001k}{2} \right| + … + \left| 500k – \frac{1001k}{2} \right| \right) $$

$$ = 2 \times \left( \frac{999k}{2} + \frac{997k}{2} + … + \frac{k}{2} \right) $$

$$ = k \times (0.5 + 1.5 + … + \text{500 terms}) \quad \text{(This is incorrect simplification, let’s stick to the AP sum)} $$
Let’s simplify the terms inside the bracket. It is a sum of odd multiples of $\frac{k}{2}$.
The terms are $\frac{k}{2}(1, 3, 5, …, 999)$.
This is an AP with first term $a = \frac{k}{2}$, common difference $d = k$, and number of terms $n_{half} = 500$.

Sum of first $N$ odd numbers is $N^2$.
$$ \sum = 2 \times \left[ \frac{k}{2} (1 + 3 + 5 + … + \text{500 terms}) \right] $$

$$ \sum = 2 \times \frac{k}{2} \times (500)^2 = k(250000) $$

Now, substitute into the Mean Deviation formula:
$$ M.D. = \frac{250000k}{1000} = 250k $$

Given $M.D. = 500$:
$$ 250k = 500 $$

$$ k = 2 $$

The question asks for $k^2$:
$$ k^2 = 2^2 = 4 $$

Ans. (2)