Statement-1 :Variance of first n natural number is `(n^2-1)/12`.
Statement-2 : S.D. of first n natural number is `sqrt((n^2-1)/12)`

Answer

To solve the problem, we need to verify the two statements regarding the variance and standard deviation of the first n natural numbers.

### Step 1: Understanding Variance
The variance \( \sigma^2 \) of a set of numbers is calculated using the formula:
\[ \sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - \left(\frac{\sum_{i=1}^{n} x_i}{n}\right)^2 \]
where \( x_i \) are the numbers in the set.

### Step 2: Calculate the Sum of the First n Natural Numbers
The first n natural numbers are \( 1, 2, 3, \ldots, n \).
The sum of the first n natural numbers is given by:
\[ \sum_{i=1}^{n} x_i = \frac{n(n+1)}{2} \]

### Step 3: Calculate the Sum of the Squares of the First n Natural Numbers
The sum of the squares of the first n natural numbers is given by:
\[ \sum_{i=1}^{n} x_i^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Substitute into the Variance Formula
Now, substituting these sums into the variance formula:
\[ \sigma^2 = \frac{\frac{n(n+1)(2n+1)}{6}}{n} - \left(\frac{\frac{n(n+1)}{2}}{n}\right)^2 \]
This simplifies to:
\[ \sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{(n+1)}{2}\right)^2 \]

### Step 5: Simplifying the Variance Expression
Now we simplify:
1. The first term is \( \frac{(n+1)(2n+1)}{6} \).
2. The second term is \( \frac{(n+1)^2}{4} \).

To combine these, we need a common denominator:
\[ \sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{3(n+1)^2}{12} \]
\[ = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12} \]
Factoring out \( (n+1) \):
\[ = \frac{(n+1)(2(2n+1) - 3(n+1))}{12} \]
Expanding the terms:
\[ = \frac{(n+1)(4n + 2 - 3n - 3)}{12} \]
\[ = \frac{(n+1)(n - 1)}{12} \]
Thus, we find:
\[ \sigma^2 = \frac{n^2 - 1}{12} \]

### Step 6: Conclusion for Statement 1
The first statement is true: the variance of the first n natural numbers is indeed \( \frac{n^2 - 1}{12} \).

### Step 7: Calculate Standard Deviation
The standard deviation \( \sigma \) is the square root of the variance:
\[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{n^2 - 1}{12}} \]

### Step 8: Conclusion for Statement 2
The second statement is also true: the standard deviation of the first n natural numbers is \( \sqrt{\frac{n^2 - 1}{12}} \).

### Final Answer
Both statements are true.

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