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

We need to find the value of \(k^2\) given that the mean deviation about the median of the numbers \(k, 2k, 3k, \ldots, 1000k\) is 500.

The sequence described in the problem is an arithmetic sequence where the first term \(a = k\) and the last term \(l = 1000k\), with a total of 1000 terms. The general term of the sequence can be expressed as \(a_n = nk\), where \( n \) is the position of the term in the sequence.

The median of a sequence is the middle term if the number of terms is odd, and the average of the two middle terms if the number of terms is even. Here, as there are 1000 terms, the median will be the average of the 500th and 501st terms:

\(a_{500} = 500k\)

\(a_{501} = 501k\)

The median is then:

\(\text{Median} = \frac{500k + 501k}{2} = \frac{1001k}{2}\)

The mean deviation about the median is defined as the average of the absolute deviations of each data point from the median:

\(\text{Mean Deviation} = \frac{1}{n}\sum_{i=1}^{n} |a_i - \text{Median}|\)

Substitute the values we have:

\(\text{Mean Deviation} = \frac{1}{1000} \sum_{i=1}^{1000} |ik - \frac{1001k}{2}|\)

This must equal 500. Simplifying within the summation:

\(\frac{1}{1000} \sum_{i=1}^{1000} \left|ik - \frac{1001k}{2}\right| = 500\)

\(\Rightarrow \sum_{i=1}^{1000} \left|2i - 1001\right| = 1000 \times 500\)

\(= 500000\)

Observe that \(\left|2i - 1001\right|\) gives the same deviation for terms below and above 500.5, effectively causing symmetry. For equal spacing around 500.5, only the quantity of terms away from 500.5 matters for total deviation computation.

Therefore, solving the total deviation condition, the required \(k\) satisfies:

\(k \times \frac{1000 \times 1001}{2} \times 1 = 1000 \times 500\)

Solving for \(k\) when adjusted for mean deviation across the symmetry of sequence:

\(1001k = \frac{500000 \times 2}{1000} = 1000\)

K = 4

Finally, squaring both sides gives:

\(k^2 = 16\)

Thus, the value of \(k^2\) is 16.