To solve the problem, we need to find the values of \( \lambda \) and \( \mu \) based on the variances given for the first \( \lambda \) natural numbers and the first \( \mu \) even natural numbers.
### Step 1: Variance of the first \( \lambda \) natural numbers
The formula for the variance \( \sigma^2 \) of the first \( n \) natural numbers is given by:
\[
\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left(\frac{n(n+1)}{2n}\right)^2
\]
This simplifies to:
\[
\sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2
\]
Given that the variance of the first \( \lambda \) natural numbers is 10, we have:
\[
\frac{(\lambda+1)(2\lambda+1)}{6} - \left(\frac{\lambda+1}{2}\right)^2 = 10
\]
### Step 2: Simplifying the equation
Now, let's simplify the equation:
1. The first term is \( \frac{(\lambda+1)(2\lambda+1)}{6} \).
2. The second term can be expanded as \( \frac{(\lambda+1)^2}{4} \).
Thus, we rewrite the equation:
\[
\frac{(\lambda+1)(2\lambda+1)}{6} - \frac{(\lambda+1)^2}{4} = 10
\]
### Step 3: Finding a common denominator
To solve this, we can find a common denominator, which is 12:
\[
\frac{2(\lambda+1)(2\lambda+1)}{12} - \frac{3(\lambda+1)^2}{12} = 10
\]
Combining the fractions gives:
\[
\frac{2(\lambda+1)(2\lambda+1) - 3(\lambda+1)^2}{12} = 10
\]
### Step 4: Cross-multiplying
Cross-multiplying gives:
\[
2(\lambda+1)(2\lambda+1) - 3(\lambda+1)^2 = 120
\]
### Step 5: Expanding and simplifying
Expanding both sides:
\[
2(2\lambda^2 + 3\lambda + 1) - 3(\lambda^2 + 2\lambda + 1) = 120
\]
This simplifies to:
\[
4\lambda^2 + 6\lambda + 2 - 3\lambda^2 - 6\lambda - 3 = 120
\]
Combining like terms gives:
\[
\lambda^2 - 1 = 120
\]
### Step 6: Solving for \( \lambda \)
Rearranging gives:
\[
\lambda^2 = 121 \implies \lambda = 11
\]
### Step 7: Variance of the first \( \mu \) even natural numbers
The first \( \mu \) even natural numbers are \( 2, 4, 6, \ldots, 2\mu \). The variance of these numbers can be expressed as:
\[
\sigma^2 = 4 \cdot \text{variance of } (1, 2, \ldots, \mu)
\]
Given that this variance is 16, we have:
\[
4 \cdot \text{variance of } (1, 2, \ldots, \mu) = 16 \implies \text{variance of } (1, 2, \ldots, \mu) = 4
\]
### Step 8: Setting up the equation for \( \mu \)
Using the variance formula again:
\[
\frac{(\mu+1)(2\mu+1)}{6} - \left(\frac{\mu+1}{2}\right)^2 = 4
\]
### Step 9: Simplifying the equation for \( \mu \)
Following similar steps as before, we find:
\[
\frac{(\mu+1)(2\mu+1)}{6} - \frac{(\mu+1)^2}{4} = 4
\]
Finding a common denominator and simplifying leads us to:
\[
\mu^2 - 1 = 48 \implies \mu^2 = 49 \implies \mu = 7
\]
### Step 10: Finding \( \lambda + \mu \)
Finally, we find:
\[
\lambda + \mu = 11 + 7 = 18
\]
### Final Answer
Thus, the value of \( \lambda + \mu \) is \( \boxed{18} \).
