If variance of first n natural number is 10 and variance of first m even natural number is 16 then the value of m+n is

Answer

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 \( m \) even natural numbers.

### Step 1: Variance of the first \( n \) natural numbers
The variance \( \sigma^2 \) of the first \( n \) natural numbers can be calculated using the formula:

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

Where:
- The sum of the first \( n \) natural numbers is given by \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \).
- The sum of the squares of the first \( n \) natural numbers is given by \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \).

Given that the variance is 10, we can set up the equation:

\[ 10 = \frac{\frac{n(n+1)(2n+1)}{6}}{n} - \left(\frac{\frac{n(n+1)}{2}}{n}\right)^2 \]

### Step 2: Simplifying the equation
This simplifies to:

\[ 10 = \frac{(n+1)(2n+1)}{6} - \left(\frac{(n+1)}{2}\right)^2 \]

Now, substituting \( \left(\frac{(n+1)}{2}\right)^2 = \frac{(n+1)^2}{4} \):

\[ 10 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \]

### Step 3: Finding a common denominator
The common denominator for 6 and 4 is 12. Thus, we rewrite the equation:

\[ 10 = \frac{2(n+1)(2n+1)}{12} - \frac{3(n+1)^2}{12} \]

Combining the fractions:

\[ 10 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12} \]

### Step 4: Clearing the fraction
Multiply both sides by 12:

\[ 120 = 2(n+1)(2n+1) - 3(n+1)^2 \]

### Step 5: Expanding and simplifying
Expanding both sides:

\[ 120 = 2(n+1)(2n+1) - 3(n^2 + 2n + 1) \]
\[ = 4n^2 + 6n + 2 - 3n^2 - 6n - 3 \]
\[ = n^2 - 1 \]

So we have:

\[ n^2 - 1 = 120 \implies n^2 = 121 \implies n = 11 \]

### Step 6: Variance of the first \( m \) even natural numbers
The first \( m \) even natural numbers are \( 2, 4, 6, \ldots, 2m \). The variance can be calculated similarly:

\[ \sigma^2 = \frac{\sum_{i=1}^{m} (2i)^2}{m} - \left(\frac{\sum_{i=1}^{m} (2i)}{m}\right)^2 \]

The sum of the first \( m \) even numbers is \( \sum_{i=1}^{m} (2i) = 2 \cdot \frac{m(m+1)}{2} = m(m+1) \).

The sum of the squares of the first \( m \) even numbers is \( \sum_{i=1}^{m} (2i)^2 = 4 \cdot \frac{m(m+1)(2m+1)}{6} = \frac{2m(m+1)(2m+1)}{3} \).

Setting the variance to 16:

\[ 16 = \frac{\frac{2m(m+1)(2m+1)}{3}}{m} - \left(\frac{m(m+1)}{m}\right)^2 \]

### Step 7: Simplifying the equation
This simplifies to:

\[ 16 = \frac{2(m+1)(2m+1)}{3} - (m+1)^2 \]

### Step 8: Finding a common denominator
Again, using a common denominator of 3:

\[ 16 = \frac{2(m+1)(2m+1) - 3(m+1)^2}{3} \]

### Step 9: Clearing the fraction
Multiply both sides by 3:

\[ 48 = 2(m+1)(2m+1) - 3(m+1)^2 \]

### Step 10: Expanding and simplifying
Expanding both sides:

\[ 48 = 2(m+1)(2m+1) - 3(m^2 + 2m + 1) \]
\[ = 4m^2 + 6m + 2 - 3m^2 - 6m - 3 \]
\[ = m^2 - 1 \]

So we have:

\[ m^2 - 1 = 48 \implies m^2 = 49 \implies m = 7 \]

### Step 11: Finding \( m+n \)
Finally, we find \( m+n \):

\[ m+n = 7 + 11 = 18 \]

### Final Answer
Thus, the value of \( m+n \) is \( \boxed{18} \).