If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16, n +

Q: Wait, why is the variance of the first $n$ natural numbers $\frac{n^2 - 1}{12}$? Can't I just calculate it manually?

You absolutely could calculate it manually for small $n$, but for JEE, time is your most precious resource! The variance of the first $n$ natural numbers is a standard result derived from $\frac{\sum x^2}{n} - (\bar{x})^2$. Memorizing this formula, $\frac{n^2 - 1}{12}$, is a high-yield strategy that saves you minutes of tedious calculation.

Q: Why does the variance become $4 \times Var(X)$ when we multiply by 2? Shouldn't it just be $2 \times Var(X)$?

That is a classic trap! Remember that variance is a measure of the *square* of the deviations. When you multiply every observation by a constant $k$, the deviations also get multiplied by $k$, and their squares get multiplied by $k^2$. Thus, $Var(kX) = k^2 Var(X)$. Since $k=2$, the variance is multiplied by $2^2 = 4$.

Q: Is there a case where $n$ or $m$ could be negative? The math gives $n^2 = 121$, so $n = \pm 11$.

Great catch on the algebra! However, in statistics, $n$ and $m$ represent the *count* of natural numbers. A count cannot be negative. Therefore, we always reject the negative root in these physical and statistical contexts. Always keep the physical meaning of your variables in mind!

Question

If variance of first $n$ natural numbers is 10 and variance of first $m$ even natural numbers is 16, $n + m$ is equal to

Sigma Percentile · Accepted Answer

### The Elegance of Statistical Patterns

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful problem in statistics. Often, students look at variance and see a daunting, complex formula involving summations and squares.

But in the world of JEE, variance is not just a formula; it is a measure of the 'spread' or 'dispersion' of data. When we look at the first $n$ natural numbers, we are looking at a perfectly uniform distribution. Let's see how we can master this.

### Phase 1: The First $n$ Natural Numbers

Imagine the set $X = \{1, 2, 3, \dots, n\}$. These numbers are perfectly spaced on the number line. The mean of this set is $\bar{x} = \frac{n+1}{2}$.

The variance, which measures how far these numbers deviate from the mean, is given by the elegant formula:

$$Var(X) = \frac{n^2 - 1}{12}$$

Why this formula? It comes from the identities:

$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \quad 	ext{and} \quad \sum_{i=1}^n i = \frac{n(n+1)}{2}$$

When you plug these into the variance definition, the algebra simplifies beautifully to $\frac{n^2 - 1}{12}$. The problem tells us this variance is $10$. So, we write:

$$\frac{n^2 - 1}{12} = 10$$

Multiplying by $12$, we get $n^2 - 1 = 120$, which leads us to $n^2 = 121$. Since $n$ must be a positive count, we find **$n = 11$**. We have cracked the first code!

### Phase 2: The Scaling Transformation

Now, let's look at the second set: the first $m$ even natural numbers, $Y = \{2, 4, 6, \dots, 2m\}$. Notice something? Every element in $Y$ is exactly twice the corresponding element in the set of the first $m$ natural numbers.

This is a linear transformation! There is a golden rule in statistics: if you scale a dataset by a constant $k$, the variance scales by $k^2$. That is:

$$Var(kX) = k^2 Var(X)$$

Here, our constant $k$ is $2$. So, the variance of our even numbers is $2^2 = 4$ times the variance of the first $m$ natural numbers. The problem tells us this variance is $16$.

So, we set up our second equation:

$$4 \cdot \frac{m^2 - 1}{12} = 16$$

Simplifying this, we divide by $4$ to get $\frac{m^2 - 1}{12} = 4$. Multiplying by $12$, we get $m^2 - 1 = 48$, which means $m^2 = 49$. Again, taking the positive root, we find **$m = 7$**.

### Phase 3: The Final Victory

We have navigated the statistical landscape and found our two unknowns: $n = 11$ and $m = 7$. The question asks for the sum $n + m$.

It is a simple final step:

$$n + m = 11 + 7 = 18$$

And there you have it! By understanding the geometric nature of variance and the power of the scaling property, we turned a potentially messy calculation into a swift, elegant solution. Keep practicing these patterns—they are the building blocks of your success in JEE Advanced! The final answer is **$18$**.

Animated Solution for Mathematics - Statistics: If variance of first $n$ natural numbers is 10 and variance of first $m$ even natural numbers is 16, $n + m$ is equal to

Enter Numerical Value:

Visualized Solution

Problem Statement

Visualizing Natural Numbers

Variance Formula for Natural Numbers

Solving for $n$ : Algebraic Steps

Finding the Value of $n$

Visualizing Even Natural Numbers

Scaling Property of Variance

Variance Formula for Even Numbers

Solving for $m$ : Algebraic Steps

Finding the Value of $m$

Final Calculation: $n + m$

The Sigma Insight: Variance and Standard Deviation

The Elegance of Statistical Patterns

Phase 1

Phase 2

Phase 3

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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 variance of first n natural numbers is 10 and variance of first m even natural numbers is 16, n+m is equal to