Find the sum of first n natural numbers.

Explanation

To find the sum of the first \( n \) natural numbers, we can use the formula for the sum of an arithmetic progression (AP). The first \( n \) natural numbers are \( 1, 2, 3, \ldots, n \).

### Step-by-Step Solution:

1. **Identify the first term (a) and the common difference (d)**:
- The first term \( a = 1 \)
- The common difference \( d = 1 \) (since each number increases by 1)

2. **Determine the number of terms (n)**:
- We want to find the sum of the first \( n \) natural numbers, so the number of terms \( n \) is simply \( n \).

3. **Use the formula for the sum of the first n terms of an arithmetic progression**:
- The formula for the sum \( S_n \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

4. **Substitute the values into the formula**:
- Plugging in the values \( a = 1 \) and \( d = 1 \):
\[ S_n = \frac{n}{2} \times (2 \times 1 + (n-1) \times 1) \]

5. **Simplify the expression**:
- This simplifies to:
\[ S_n = \frac{n}{2} \times (2 + n - 1) \]
- Which further simplifies to:
\[ S_n = \frac{n}{2} \times (n + 1) \]

6. **Final result**:
- Therefore, the sum of the first \( n \) natural numbers is:
\[ S_n = \frac{n(n + 1)}{2} \]