lf the mean deviation about the median of the numbers k, 2k, 3k, ..... , 1000k is 500, then $$k^{2}$$ is equal to :

We have the data set $$k,2k,3k,\dots ,1000k$$; that is an arithmetic progression with first term $$k$$ and common difference $$k$$.

Total number of terms, $$N=1000$$, is even. For an even-sized ordered list, the median is the average of the $$\frac{N}{2}$$-th and $$\left(\frac{N}{2}+1\right)$$-th terms.

The $$500$$-th term is $$500k$$ and the $$501$$-st term is $$501k$$.
Hence the median, $$M$$, equals $$\dfrac{500k+501k}{2}=\dfrac{1001k}{2}=500.5k$$.

Mean deviation about the median is defined as
$$\text{M.D._{(median)}}=\frac{1}{N}\sum_{i=1}^{N}\left|x_i-M\right|.$$

Here
$$\text{M.D.}=\frac{1}{1000}\sum_{i=1}^{1000}\left|ik-500.5k\right|$$ $$=\frac{k}{1000}\sum_{i=1}^{1000}\left|i-500.5\right|.$$

Let
$$S=\sum_{i=1}^{1000}\left|i-500.5\right|.$$ Because the list is symmetric around $$500.5$$, the absolute differences come in equal pairs.

For $$i=1,2,\dots ,500$$ we have distances
$$499.5,498.5,\dots ,0.5,$$ an arithmetic progression with
first term $$a_1=499.5$$, last term $$a_{500}=0.5$$, number of terms $$n=500$$.
Sum of these 500 terms:
$$S_{500}=\frac{n}{2}\left(a_1+a_{500}\right)=\frac{500}{2}(499.5+0.5)=250\times500=125\,000.$$

The same sum occurs for $$i=501,502,\dots ,1000$$ (distances $$0.5,1.5,\dots ,499.5$$).
Therefore
$$S=2\times125\,000=250\,000.$$

Given in the question, mean deviation about the median equals $$500$$, so
$$\frac{k}{1000}\times250\,000=500.$$

Simplifying:
$$k\times250=500 \quad\Longrightarrow\quad k=2.$$

Hence
$$k^{2}=2^{2}=4.$$

The value of $$k^{2}$$ is $$4$$, which corresponds to Option C.