If the variance of first n natural numbers is 2 and the variance of first m odd natural numbers is 40, then `m +n =`

Explanation

To solve the problem, we need to find the values of \( n \) and \( m \) based on the given variances of the first \( n \) natural numbers and the first \( m \) odd natural numbers.

### Step 1: Variance of the first \( n \) natural numbers
The formula for the variance \( \sigma^2 \) of the first \( n \) natural numbers is given by:
\[ \sigma^2 = \frac{n^2 - 1}{12} \]
According to the problem, this variance is equal to 2:
\[ \frac{n^2 - 1}{12} = 2 \]

### Step 2: Solve for \( n \)
To solve for \( n \), we first multiply both sides of the equation by 12:
\[ n^2 - 1 = 24 \]
Next, we add 1 to both sides:
\[ n^2 = 25 \]
Taking the square root of both sides, we find:
\[ n = 5 \quad (\text{since } n \text{ must be positive}) \]

### Step 3: Variance of the first \( m \) odd natural numbers
The formula for the variance of the first \( m \) odd natural numbers is:
\[ \sigma^2 = \frac{4m^2 - 1}{12} \]
According to the problem, this variance is equal to 40:
\[ \frac{4m^2 - 1}{12} = 40 \]

### Step 4: Solve for \( m \)
To solve for \( m \), we first multiply both sides of the equation by 12:
\[ 4m^2 - 1 = 480 \]
Next, we add 1 to both sides:
\[ 4m^2 = 481 \]
Now, we divide both sides by 4:
\[ m^2 = \frac{481}{4} \]
Taking the square root of both sides, we find:
\[ m = \frac{\sqrt{481}}{2} \]
Calculating \( \sqrt{481} \) gives us approximately \( 22 \), thus:
\[ m \approx \frac{22}{2} = 11 \]

### Step 5: Calculate \( m + n \)
Now that we have \( n = 5 \) and \( m = 11 \), we can find \( m + n \):
\[ m + n = 11 + 5 = 16 \]

### Final Answer:
\[ m + n = 16 \]