Step 1: Test both statements with one concrete example, instead of proving them in general first.
Take a small, easy case: let $n = 5$, so the first 5 natural numbers are $1, 2, 3, 4, 5$.
Step 2: Check Reason (R) with this example.
The actual sum is $1+2+3+4+5 = 15$. The formula in Reason (R) gives:
\[ \frac{n(n+1)}{2} = \frac{5 \times 6}{2} = 15 \]
This matches the actual sum exactly, so Reason (R) checks out as true for this case, and since it is a standard general identity, it is true for every $n$.
Step 3: Check Assertion (A) with the same example.
The actual mean of $1, 2, 3, 4, 5$ is:
\[ \text{Mean} = \frac{15}{5} = 3 \]
Assertion (A) claims the mean should be $\frac{n-1}{2}$. For $n = 5$, this gives:
\[ \frac{n-1}{2} = \frac{5-1}{2} = 2 \]
But the real mean we just calculated is 3, not 2. Since $3 \neq 2$, Assertion (A) does not hold for this example, so Assertion (A) is false.
Step 4: Note the correct formula for reference.
The correct mean of the first $n$ natural numbers is actually $\frac{n+1}{2}$ (sum divided by count), which for $n=5$ gives $\frac{6}{2}=3$, matching our direct calculation.
Step 5: Final Answer.
Assertion (A) is false, but Reason (R) is true, so option (D) is correct.
\[ \boxed{\text{A is false, R is true}} \]