Question

If the mean deviation about the median of the numbers \[ k,\,2k,\,3k,\,\ldots,\,1000k \] is \(500\), then \(k^2\) is equal to

• A. \(4\)
• B. \(16\)
• C. \(1\)
• D. \(9\)

Hint:

For equally spaced symmetric data, mean deviation about the median depends directly on the common difference.

Answer 1

The Correct Option is B

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, \[ k^2 = 16. \]

Final Answer: \[ \boxed{16} \]