Question

Mean deviation about median for \( k, 2k, 3k, \ldots, 1000k \) is 500, then the value of \( k^2 \) is:

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

Hint:

For symmetric data:
Median lies at the center
Mean deviation can be calculated efficiently using symmetry

Answer 1

The Correct Option is A

Given:
The data is \(k, 2k, 3k, \ldots, 1000k\)
Mean deviation about median = 500

Step 1: Find the median

Total number of terms = 1000 (even). The median is the average of the 500th and 501st terms.

\[ \text{Median} = \frac{500k + 501k}{2} = \frac{1001k}{2} \]

Step 2: Formula for mean deviation about median

\[ \text{MD} = \frac{1}{n}\sum |x_i - \text{Median}| \]

Here, \(n = 1000\)

Step 3: Absolute deviations

Deviations are symmetric about the median.

\[ |rk - \tfrac{1001k}{2}| = k\left|\tfrac{1001}{2} - r\right| \]

Hence,

\[ \sum |x_i - \text{Median}| = k \sum_{r=1}^{1000} \left|\tfrac{1001}{2} - r\right| \]

Step 4: Evaluate the sum

The sum of absolute deviations for first 1000 natural numbers about their median is:

\[ \sum_{r=1}^{1000} \left|\tfrac{1001}{2} - r\right| = 250000 \]

Step 5: Mean deviation

\[ \text{MD} = \frac{1}{1000} \times k \times 250000 \]

\[ \text{MD} = 250k \]

Given mean deviation = 500

\[ 250k = 500 \]

\[ k = 2 \]

Final Answer:

\[ k^2 = 4 \]