JEE Main 2026 (22 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Statistics: If the mean deviation about the median of the numbers $k, 2 k, 3 k, \dots, 1000 k$ is 500, then $k^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Sequence

Sequence: $k, 2 k, 3 k, \dots, 1000 k$
Total terms, $n = 1000$ (Even)
This is an Arithmetic Progression (A.P.)

Finding the Median Position

For even $n$ , Median is the average of middle two terms.
Middle terms: $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$
Here, $50 0^{th}$ and $50 1^{st}$ terms.

Calculating the Median

$50 0^{th}$ term = $500 k$
$50 1^{st}$ term = $501 k$
Median $M = \frac{500 k + 501 k}{2} = 500.5 k$

Mean Deviation Formula

Mean Deviation ( $M D$ ) = $\frac{1}{n} \sum_{i = 1}^{n} ∣ x_{i} - M ∣$
It measures the average distance of all points from the median.

Setting up the Mean Deviation

Substitute $x_{i} = ik$ and $M = 500.5 k$
$M D = \frac{1}{1000} \sum_{i = 1}^{1000} ∣ ik - 500.5 k ∣$
Factor out $k$ : $M D = \frac{k}{1000} \sum_{i = 1}^{1000} ∣ i - 500.5∣$

Exploiting Symmetry

The distances are symmetric about $500.5$ .
Distance of $1^{st}$ term = Distance of $100 0^{th}$ term.
$\sum_{i = 1}^{1000} ∣ i - 500.5∣ = 2 \sum_{i = 1}^{500} (500.5 - i)$

Expanding the Summation

Let $S = \sum_{i = 1}^{500} (500.5 - i)$
Split the sum: $S = \sum_{i = 1}^{500} 500.5 - \sum_{i = 1}^{500} i$
$S = (500 \times 500.5) - \frac{500 \times 501}{2}$

Calculating the Sum

$500 \times 500.5 = 250250$
$\frac{500 \times 501}{2} = 250 \times 501 = 125250$
$S = 250250 - 125250 = 125000$

Total Sum of Deviations

Total Sum = $2 \times S$
Total Sum = $2 \times 125000 = 250000$
Substitute back into $M D$ :
$M D = \frac{k}{1000} \times 250000$

Simplifying Mean Deviation

$M D = \frac{250000}{1000} \times k$
$M D = 250 k$

Solving for $k$

We are given that $M D = 500$ .
Equating the two: $250 k = 500$
$k = \frac{500}{250} = 2$

Final Answer: $k^{2}$

The question asks for the value of $k^{2}$ .
$k^{2} = 2^{2} = 4$
Final Answer: 4

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The Sigma Insight: Measures of Dispersion

Solution Diagram

Analyzing the Setup

Imagine standing on a vast, perfectly flat plain. You have 1000 markers placed at intervals of

k

This is not just a sequence; it is an Arithmetic Progression, a rhythmic heartbeat of numbers:

k, 2 k, 3 k, \dots, 1000 k

In the world of JEE, recognizing the structure of a problem is half the battle won.

The Median Trap

In statistics, the median is the heart of the data. However, we face a classic JEE trap here because we have 1000 terms, which is an even number.

There is no single middle term. We must find the average of the two middlemost terms, which are the

50 0^{th}

term (

500 k

) and the

50 1^{st}

term (

501 k

Their average is our pivot point, the axis of symmetry:

M = \frac{500 k + 501 k}{2} = 500.5 k

The Mean Deviation

Now, we calculate the Mean Deviation (

M D

), which is the average distance of all points from this median.

Mathematically, this is expressed as:

M D = \frac{1}{1000} i = 1 \sum 1000 ∣ ik - 500.5 k ∣

Factoring out the constant

k

, we simplify the expression to:

M D = \frac{k}{1000} i = 1 \sum 1000 ∣ i - 500.5∣

The Power of Symmetry

Do not calculate 1000 absolute values individually. Look at the sequence: the distance of the

1^{st}

term (

1 k

) from

500.5 k

is identical to the distance of the

100 0^{th}

term (

1000 k

) from

500.5 k

By exploiting this symmetry, we can calculate the sum for the first 500 terms and simply double it:

M D = \frac{k}{1000} \times 2 i = 1 \sum 500 (500.5 - i)

The Final Calculation

Expanding the summation, we get:

2 \times [(500 \times 500.5) - i = 1 \sum 500 i]

Using the sum of natural numbers formula,

\frac{n ( n + 1 )}{2}

, the sum of the first 500 natural numbers is:

\frac{500 \times 501}{2} = 125, 250

The first part of the expression is

500 \times 500.5 = 250, 250

. Subtracting these values gives

125, 000

, and multiplying by 2 yields

250, 000

Thus, the Mean Deviation is:

M D = \frac{k}{1000} \times 250, 000 = 250 k

Given

M D = 500

, we solve

250 k = 500

to find

k = 2

. The question asks for

k^{2}

, therefore:

Result = 4

Common Traps & FAQs

Wait, why is the median

500.5 k

and not just

500 k

That is a fantastic question! In a sequence with an even number of terms, there is no single middle term. We must take the average of the two central terms. Here, the

50 0^{th}

term is

500 k

and the

50 1^{st}

term is

501 k

. Their average is

\frac{500 k + 501 k}{2} = 500.5 k

. Always check if your

n

is odd or even before finding the median!

Why did we multiply the sum by 2 in the symmetry step?

Because of the perfect symmetry of an Arithmetic Progression! The distance of the

1^{st}

term from the median is the same as the distance of the

100 0^{th}

term. The distance of the

2^{nd}

term is the same as the

99 9^{th}

, and so on. Instead of calculating 1000 absolute values, we calculate the sum for the first 500 terms and simply double it. It is a massive time-saver!

Is it okay to just memorize the formula for Mean Deviation of first

n

natural numbers?

While you can memorize formulas, understanding the derivation is what makes you a JEE topper. The formula for the Mean Deviation of the first

n

natural numbers is

\frac{n ^{2} - 1}{4 n}

for odd

n

and

\frac{n}{4}

for even

n

. However, deriving it using the symmetry argument as we did here ensures you won't get confused if the sequence is slightly different. Stick to the logic!

Why does the question ask for

k^{2}

instead of just

k

This is a classic JEE trap designed to test your focus. Many students find

k = 2

and immediately rush to mark it as the answer, forgetting to read the final requirement. Always, and I mean always, re-read the question statement before bubbling your answer. It asks for

k^{2}

, so

2^{2} = 4

is the final step.

Animated Solution for Mathematics - Statistics: If the mean deviation about the median of the numbers $k, 2 k, 3 k, \dots, 1000 k$ is 500, then $k^{2}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Sequence

Finding the Median Position

Calculating the Median

Mean Deviation Formula

Setting up the Mean Deviation

Exploiting Symmetry

Expanding the Summation

Calculating the Sum

Total Sum of Deviations

Simplifying Mean Deviation

Solving for $k$

Final Answer: $k^{2}$

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Median Trap

The Mean Deviation

The Power of Symmetry

The Final Calculation

Similar Questions

If the mean deviation about the mean of the numbers $1, 2, 3, \dots, n$ , where $n$ is odd, is $\frac{5 ( n + 1 )}{n}$ , then $n$ is equal to ______.

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Animated Solution for Mathematics - Statistics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean deviation about the median of the numbers k,2k,3k,…,1000k is 500, then k2 is equal to :

Select Answer:

Visualized Solution

Visualizing the Sequence

Finding the Median Position

Calculating the Median

Mean Deviation Formula

Setting up the Mean Deviation

Exploiting Symmetry

Expanding the Summation

Calculating the Sum

Total Sum of Deviations

Simplifying Mean Deviation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k2

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Median Trap

The Mean Deviation

The Power of Symmetry

The Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the mean of 6 observations 1, 2, 4, 5, x and y be 5 and their variance be 10. Then their mean deviation about the mean is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The variance σ2 of the data xi​fi​​03​12​53​62​106​123​173​​ is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x1​,x2​,…,xn​ be n observations such that ∑xi2​=400 and ∑xi​=80. Then the possible value of n among the following is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and 437​ respectively, then (a−b)2 is equal to:

Animated Solution for Mathematics - Statistics: If the mean deviation about the median of the numbers $k, 2 k, 3 k, \dots, 1000 k$ is 500, then $k^{2}$ is equal to :

Solving for $k$

Final Answer: $k^{2}$

If the mean deviation about the mean of the numbers $1, 2, 3, \dots, n$ , where $n$ is odd, is $\frac{5 ( n + 1 )}{n}$ , then $n$ is equal to ______.

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and their variance is $6.8$ . If $M$ is the mean deviation of the numbers about the mean, then $25 M$ is equal to:

Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to

The variance $σ^{2}$ of the data $x_{i} f_{i} 03125362106123173$ is ________.

Let $x_{1}, x_{2}, \dots, x_{n}$ be n observations such that $\sum x_{i}^{2} = 400$ and $\sum x_{i} = 80$ . Then the possible value of n among the following is

If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and $\frac{37}{4}$ respectively, then $(a - b)^{2}$ is equal to: