The mean of the first $n$ natural numbers is given by the formula:

$\text{Mean} = \frac{\text{Sum of the first } n \text{ natural numbers}}{n}$

The sum of the first $n$ natural numbers is:

$S_n = \frac{n(n+1)}{2}$

So the mean is:

$\text{Mean} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$

According to the problem, this mean is also given as $\frac{5n}{9}$. Therefore, we can set up the equation:

$\frac{n+1}{2} = \frac{5n}{9}$

Now solve for $n$.

1. Multiply both sides by 18 to eliminate the denominators:

$9(n+1) = 10n$

2. Expand and simplify:

$9n + 9 = 10n$

$9 = 10n - 9n$

$n = 9$

Thus, $n = 9$.

Would you like further details on the steps or have any questions?

Here are some related questions to deepen understanding:

1. How do you calculate the sum of the first $n$ natural numbers?
2. What is the formula for the mean of a sequence of numbers?
3. Can you generalize the formula for the mean of the first $n$ even numbers?
4. What would happen if the mean was $\frac{6n}{10}$ instead of $\frac{5n}{9}$?
5. How can the arithmetic mean be used to solve other types of problems involving sequences?

Tip: In any arithmetic sequence, the mean is equal to the average of the first and last term!