Let n be an odd natural number such that the variance of 1, 2, 3, 4, ..., n is 14. Then n is equal t

Q: Wait, why do we use the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ instead of the definition $\frac{\sum (x_i - \bar{x})^2}{n}$?

Both are mathematically equivalent! However, the formula $\frac{\sum x_i^2}{n} - (\bar{x})^2$ is often much faster for sequences like this because it avoids calculating the deviation of every single term from the mean. It turns a summation of differences into a simple algebraic manipulation.

Q: Why do we reject the negative root when solving $n^2 = 169$?

Great catch! In the context of this problem, $n$ represents the number of terms in a sequence. A sequence cannot have a negative number of terms. Since $n$ must be a natural number, we only accept the positive root, $n = 13$.

Q: Is there a shortcut to remember the variance of the first $n$ natural numbers?

Yes, it is a very useful result for JEE Advanced! The variance of the first $n$ natural numbers is always $\frac{n^2-1}{12}$. Memorizing this can save you precious minutes during the exam, but always be ready to derive it if you forget!

Question

Let $n$ be an odd natural number such that the variance of $1, 2, 3, 4, \dots, n$ is 14. Then $n$ is equal to .

Sigma Percentile · Accepted Answer

### The Elegance of Dispersion: Unlocking the Variance of Natural Numbers

Imagine you are standing on a number line, looking at the first $n$ natural numbers: $1, 2, 3, \dots, n$. These numbers are perfectly spaced, like rungs on a ladder.

The problem asks us to find $n$, given that the variance of this sequence is exactly $14$. This isn't just a dry statistical calculation; it is a journey into the heart of how data spreads around its center.

### Phase 1: The Statistical Toolkit

To solve this, we need to define our tools. Variance, denoted by $\sigma^2$, measures the average squared deviation from the mean.

While the definition $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$ is conceptually beautiful, it is often computationally heavy. Instead, we use the computational formula:

$$\sigma^2 = \frac{\sum_{i=1}^n x_i^2}{n} - \left(\frac{\sum_{i=1}^n x_i}{n}\right)^2$$

This formula is our secret weapon. It separates the problem into two manageable parts: the mean of the squares and the square of the mean.

### Phase 2: Calculating the Mean and the Mean of Squares

First, let's find the mean, $\bar{x}$. The sum of the first $n$ natural numbers is a classic result: $\sum_{i=1}^n i = \frac{n(n+1)}{2}$.

To find the mean, we divide this sum by the number of terms, $n$:

$$\bar{x} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$$

Next, we need the mean of the squares. The sum of the squares of the first $n$ natural numbers is given by $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$.

Again, we divide by $n$ to get the mean of the squares:

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

### Phase 3: The Algebraic Symphony

Now, we bring these pieces together into our variance formula. Substitute our expressions back in:

$$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$$

I know this looks like a mess of fractions, but take a deep breath. Let's simplify by factoring out $\frac{n+1}{2}$:

$$\sigma^2 = \frac{n+1}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right]$$

Finding a common denominator inside the bracket gives us $\frac{2(2n+1) - 3(n+1)}{6} = \frac{4n+2 - 3n - 3}{6} = \frac{n-1}{6}$.

Multiplying this by the term outside, we get:

$$\sigma^2 = \frac{n+1}{2} \cdot \frac{n-1}{6} = \frac{n^2-1}{12}$$

This is the elegant result we were looking for! The variance of the first $n$ natural numbers is always $\frac{n^2-1}{12}$.

### Phase 4: The Final Resolution

We are almost there. The problem tells us that the variance is $14$. So, we set our derived formula equal to $14$:

$$\frac{n^2-1}{12} = 14$$

Multiply both sides by $12$:

$$n^2 - 1 = 168$$

$$n^2 = 169$$

Taking the square root, we find $n = 13$ or $n = -13$. Since $n$ must be a natural number, we reject the negative value.

Thus, **$n = 13$**. We have successfully navigated the algebra and arrived at the solution.

Animated Solution for Mathematics - Statistics: Let $n$ be an odd natural number such that the variance of $1, 2, 3, 4, \dots, n$ is 14. Then $n$ is equal to .

Enter Numerical Value:

Visualized Solution

Visualize the Sequence

General Variance Formula

Mean of First $n$ Natural Numbers

Mean of Squares

Substitute into Variance Formula

Simplify the Expression

Equate to Given Variance

Isolate $n^{2} - 1$

Solve for $n^{2}$

Find the Value of $n$

Final Verification

The Sigma Insight: Measures of Dispersion

The Elegance of Dispersion

Phase 1

Phase 2

Phase 3

Phase 4

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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; } Let n be an odd natural number such that the variance of 1,2,3,4,…,n is 14. Then n is equal to .