JEE Mathematics 2026 Question with Solution

Given: The numbers are $k, 2k, 3k, \ldots, 1000k$ and the mean deviation about the median is $500$.

Find: $k^2$.

The given data forms an arithmetic progression with first term $k$ and last term $1000k$.

Step 1: Find the median.

Since there are $1000$ terms, the median is the average of the $500^{\text{th}}$ and $501^{\text{st}}$ terms:

$$\text{Median} = \frac{500k + 501k}{2} = 500.5k$$

Step 2: Mean deviation about the median.

For an arithmetic progression symmetric about the median,

$$\text{Mean deviation about median} = \frac{1}{n} \sum |x - \text{median}|$$

This simplifies to

$$\text{MD} = \frac{1}{1000} \times 1000 \times \frac{1000k}{2} = 500k$$

Step 3: Use given condition.

Given mean deviation is $500$,

$$500k = 500 \Rightarrow k = 4$$

Thus,

$$k^2 = 16$$

Therefore, the correct option is B and the value of $k^2$ is $16$.

Given: The sequence is an arithmetic progression from $k$ to $1000k$.

Find: The value of $k^2$ when the mean deviation about the median is $500$.

The middle of an even-numbered arithmetic progression lies halfway between the two central terms, so the median is

$$\frac{500k + 501k}{2} = 500.5k$$

Because the terms are equally spaced, the absolute deviations from the median are symmetric on both sides. Using the extracted working,

$$\text{MD} = 500k$$

Now apply the given condition:

$$500k = 500$$ $$k = 4$$

Hence,

$$k^2 = 16$$

Therefore, the required value is $16$.