Median Calculation for the Sequence

The given sequence is $k, 2k, 3k, \dots, 1000k$. This is an arithmetic progression with $N=1000$ terms.

Since the number of terms ($N=1000$) is even, the median ($M$) is the average of the $\frac{N}{2}^{th}$ and $(\frac{N}{2} + 1)^{th}$ terms.

• $\frac{N}{2} = \frac{1000}{2} = 500$. The $500^{th}$ term is $500k$.
• $\frac{N}{2} + 1 = 501$. The $501^{st}$ term is $501k$.

Median $M = \frac{500k + 501k}{2} = \frac{1001k}{2} = 500.5k$.

Mean Deviation Calculation

The mean deviation (MD) about the median is calculated using the formula:

$ MD = \frac{1}{N} \sum_{i=1}^{N} |x_i - M| $

Here, $x_i = ik$ and $M = 500.5k$. The terms are:

$ |x_i - M| = |ik - 500.5k| = k |i - 500.5| $

The sum of the absolute deviations is:

$ \sum_{i=1}^{1000} |x_i - M| = \sum_{i=1}^{1000} k |i - 500.5| = k \sum_{i=1}^{1000} |i - 500.5| $

We can split the sum:

• For $i = 1$ to $500$: $|i - 500.5| = 500.5 - i$
• For $i = 501$ to $1000$: $|i - 500.5| = i - 500.5$

Sum $= k \left[ \sum_{i=1}^{500} (500.5 - i) + \sum_{i=501}^{1000} (i - 500.5) \right]$

Calculating the sums:

• $\sum_{i=1}^{500} (500.5 - i) = (500 \times 500.5) - (\sum_{i=1}^{500} i) = 250250 - \frac{500 \times 501}{2} = 250250 - 125250 = 125000$.
• $\sum_{i=501}^{1000} (i - 500.5) = (\sum_{i=501}^{1000} i) - (500 \times 500.5) = (\frac{1000 \times 1001}{2} - \frac{500 \times 501}{2}) - 250250 = (500500 - 125250) - 250250 = 375250 - 250250 = 125000$.

Total sum of absolute deviations $= k [125000 + 125000] = 250000k$.

Mean Deviation $MD = \frac{250000k}{1000} = 250k$.

Solving for k^2

We are given that the mean deviation about the median is 500.

$ MD = 250k = 500 $

Solving for $k$:

$ k = \frac{500}{250} = 2 $

We need to find $k^2$.

$ k^2 = 2^2 = 4 $