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-by-Step Solution:

#### 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:

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

The sum of the first \( n \) natural numbers is:

\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]

The sum of the squares of the first \( n \) natural numbers is:

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

Substituting these 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:

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

Calculating the second term:

\[ \left(\frac{(n+1)}{2}\right)^2 = \frac{(n+1)^2}{4} \]

Thus, the variance becomes:

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

Setting this equal to 10 (as given in the problem):

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

#### Step 2: Solve for \( n \)
To eliminate the fractions, multiply through by 12:

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

Expanding this gives:

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

This simplifies to:

\[ 4n^2 + 12n + 8 - 3n^2 - 6n - 3 = 120 \]

Combining like terms results in:

\[ n^2 + 6n + 5 = 120 \]

Rearranging gives:

\[ n^2 + 6n - 115 = 0 \]

Using the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 + 460}}{2} = \frac{-6 \pm 22}{2} \]

Calculating the roots:

\[ n = \frac{16}{2} = 8 \quad \text{(valid)} \quad \text{or} \quad n = \frac{-28}{2} = -14 \quad \text{(not valid)} \]

So, \( n = 8 \).

#### Step 3: 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:

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

Calculating the sums:

\[ \sum_{i=1}^{m} (2i) = 2 \cdot \frac{m(m+1)}{2} = m(m+1) \]

\[ \sum_{i=1}^{m} (2i)^2 = 4 \cdot \sum_{i=1}^{m} i^2 = 4 \cdot \frac{m(m+1)(2m+1)}{6} \]

Substituting into the variance formula:

\[ \sigma^2 = \frac{4 \cdot \frac{m(m+1)(2m+1)}{6}}{m} - \left(\frac{m(m+1)}{m}\right)^2 \]

This simplifies to:

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

Setting this equal to 16:

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

#### Step 4: Solve for \( m \)
Multiply through by 3 to eliminate the fraction:

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

Expanding gives:

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

This simplifies to:

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

Combining like terms results in:

\[ m^2 - 1 = 48 \]

Thus:

\[ m^2 = 49 \implies m = 7 \quad \text{(valid)} \quad \text{or} \quad m = -7 \quad \text{(not valid)} \]

So, \( m = 7 \).

#### Step 5: Calculate \( m + n \)
Now that we have \( n = 8 \) and \( m = 7 \):

\[ m + n = 7 + 8 = 15 \]

### Final Answer:
Thus, the value of \( m + n \) is \( \boxed{15} \).