If the mean deviation about the median of the numbers $k,\ 2k,\ 3k,\ \ldots,\ 1000k$ is $500$, then $k^2$ is equal to:
A) 1
B) 16
C) 9
D) 4 ✓
✓
Solution Steps
1
Identify the data set
Terms: $k,\ 2k,\ 3k,\ \ldots,\ 1000k$ (i.e. $x_i = ik$ for $i=1$ to $1000$)
Number of terms: $n = 1000$ (even)
The sequence is already sorted in ascending order (assuming $k > 0$).
2
Find the Median
For even $n$, median $= \dfrac{\text{(n/2)th term} + \text{(n/2+1)th term}}{2}$
$$M = \frac{500k + 501k}{2} = \frac{1001k}{2} = 500.5k$$
3
Set up the Mean Deviation formula
$$\text{MD} = \frac{1}{n}\sum_{i=1}^{1000}|x_i – M| = \frac{1}{1000}\sum_{i=1}^{1000}|ik – 500.5k|$$
$$= \frac{k}{1000}\sum_{i=1}^{1000}|i – 500.5|$$
4
Evaluate Σ|i − 500.5|
Split the sum at $i = 500$ and $i = 501$:
For $i = 1$ to $500$: $|i – 500.5| = 500.5 – i$
For $i = 501$ to $1000$: $|i – 500.5| = i – 500.5$
For $i = 501$ to $1000$: $|i – 500.5| = i – 500.5$
By symmetry both halves are equal:
$$\sum_{i=1}^{1000}|i-500.5| = 2\sum_{i=1}^{500}(500.5-i)$$
$$= 2\!\left[500\times500.5 – \sum_{i=1}^{500}i\right] = 2\!\left[250250 – \frac{500\times501}{2}\right]$$
$$= 2\left[250250 – 125250\right] = 2\times125000 = 250000$$
5
Compute MD and solve for k
$$\text{MD} = \frac{k}{1000}\times 250000 = 250k$$
Setting MD $= 500$:
$$250k = 500 \implies k = 2$$
$$\therefore\; k^2 = \boxed{4}$$
✓ Answer: D) 4
💡
Key Insight
Median=500.5k → Σ|i−500.5|=250000 → MD=250k=500 → k=2 → k²=4
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Theory: Statistics — All Key Formulas
Mean Deviation (MD)
About Mean ($\bar{x}$):
$$\text{MD}_{\bar{x}} = \frac{1}{n}\sum_{i=1}^{n}|x_i – \bar{x}|$$
About Median ($M$):
$$\text{MD}_{M} = \frac{1}{n}\sum_{i=1}^{n}|x_i – M|$$
Key Property: MD is minimized when taken about the median.
So $\text{MD}_M \leq \text{MD}_{\bar{x}}$ always.
So $\text{MD}_M \leq \text{MD}_{\bar{x}}$ always.
Median — How to Find
| Case | Formula |
|---|---|
| $n$ odd | $M = x_{\left(\frac{n+1}{2}\right)}$ |
| $n$ even | $M = \dfrac{x_{(n/2)} + x_{(n/2+1)}}{2}$ |
| Grouped data | $M = l + \dfrac{\frac{n}{2}-cf}{f}\times h$ |
For $k,2k,\ldots,1000k$: $n=1000$ (even), $M = \dfrac{500k+501k}{2} = 500.5k$
Variance & Standard Deviation
$$\sigma^2 = \frac{1}{n}\sum(x_i-\bar{x})^2 = \frac{\sum x_i^2}{n} – \bar{x}^2$$
$$\sigma = \sqrt{\sigma^2}$$
For AP with first term $a$, common difference $d$, $n$ terms:
$\bar{x} = a + \dfrac{(n-1)d}{2}$, $\quad \sigma^2 = \dfrac{(n^2-1)d^2}{12}$
Here: $a=k$, $d=k$, $n=1000$ → $\bar{x} = k + \dfrac{999k}{2} = \dfrac{1001k}{2} = 500.5k$ (same as median for symmetric AP!)
$\bar{x} = a + \dfrac{(n-1)d}{2}$, $\quad \sigma^2 = \dfrac{(n^2-1)d^2}{12}$
Here: $a=k$, $d=k$, $n=1000$ → $\bar{x} = k + \dfrac{999k}{2} = \dfrac{1001k}{2} = 500.5k$ (same as median for symmetric AP!)
Measures of Dispersion — Summary
| Measure | Formula | Best used when |
|---|---|---|
| Range | Max − Min | Quick estimate |
| MD about mean | $\frac{1}{n}\sum|x_i-\bar{x}|$ | Symmetric data |
| MD about median | $\frac{1}{n}\sum|x_i-M|$ | Skewed data |
| Variance $\sigma^2$ | $\frac{1}{n}\sum(x_i-\bar{x})^2$ | Mathematical use |
| SD $\sigma$ | $\sqrt{\sigma^2}$ | Same units as data |
| CV | $\frac{\sigma}{\bar{x}}\times100\%$ | Comparing datasets |
Symmetry Trick for AP
For a symmetric AP (equal no. of terms on both sides of median):
$$\sum_{i=1}^{n}|x_i – M| = 2\sum_{i=1}^{n/2}(M – x_i)$$
This halves the computation! Always use this for AP problems in JEE.
General formula: For $1,2,3,\ldots,2m$ with median $= m+0.5$:
$$\sum_{i=1}^{2m}|i-(m+0.5)| = 2\sum_{i=1}^{m}(m+0.5-i) = m^2$$
Here $m=500$: sum $= 500^2 = 250000$ ✓
$$\sum_{i=1}^{n}|x_i – M| = 2\sum_{i=1}^{n/2}(M – x_i)$$
This halves the computation! Always use this for AP problems in JEE.
General formula: For $1,2,3,\ldots,2m$ with median $= m+0.5$:
$$\sum_{i=1}^{2m}|i-(m+0.5)| = 2\sum_{i=1}^{m}(m+0.5-i) = m^2$$
Here $m=500$: sum $= 500^2 = 250000$ ✓
❓
Frequently Asked Questions
Q
Why is the median 500.5k and not 500k?
▼
n=1000 is even, so median = average of 500th and 501st terms = (500k+501k)/2 = 1001k/2 = 500.5k. If n were odd, median would be the middle term exactly.
Q
How does symmetry simplify the sum Σ|i−500.5|?
▼
The sequence 1,2,…,1000 is symmetric about 500.5. The deviation of i from below (|i−500.5| for i≤500) mirrors that from above. So total sum = 2× lower half sum = 2×125000 = 250000.
Q
How is Σᵢ₌₁⁵⁰⁰ i = 125250?
▼
Σᵢ₌₁⁵⁰⁰ i = 500×501/2 = 250500/2 = 125250. This uses the formula Σᵢ₌₁ⁿ i = n(n+1)/2.
Q
What if k is negative?
▼
If k<0, the sequence is descending. The median would still be 500.5k (negative) and MD=250|k|=500 → |k|=2 → k²=4. Same answer regardless of sign of k.
Q
Is MD about median always less than MD about mean?
▼
Yes! The median minimizes Σ|xᵢ−a| over all choices of a. So MD about median ≤ MD about mean. For symmetric distributions (like this AP), mean = median, so both are equal here.
Q
Quick formula: For AP 1,2,…,2m — what is Σ|i−(m+0.5)|?
▼
Σᵢ₌₁²ᵐ |i−(m+0.5)| = m². Here m=500, so sum=500²=250000. This is a very handy shortcut for JEE problems involving mean deviation of an AP.
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