If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p+q) terms.

Explanation

To solve the problem, we need to find the sum of the first (p + q) terms of an arithmetic progression (A.P.) given that the sum of the first p terms is equal to the sum of the first q terms.

### Step-by-Step Solution:

1. **Understanding the Sum of the First n Terms of an A.P.**:
The sum of the first n terms \( S_n \) of an A.P. can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Setting Up the Equations**:
Given that the sum of the first p terms is equal to the sum of the first q terms, we can write:
\[ S_p = S_q \]
This translates to:
\[ \frac{p}{2} \times (2a + (p - 1)d) = \frac{q}{2} \times (2a + (q - 1)d) \]

3. **Eliminating the Common Factor**:
We can multiply both sides by 2 to eliminate the fraction:
\[ p(2a + (p - 1)d) = q(2a + (q - 1)d) \]

4. **Expanding Both Sides**:
Expanding both sides gives us:
\[ 2ap + pd(p - 1) = 2aq + qd(q - 1) \]

5. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 2ap - 2aq + pd(p - 1) - qd(q - 1) = 0 \]
This can be factored as:
\[ 2a(p - q) + d \left( p(p - 1) - q(q - 1) \right) = 0 \]

6. **Factoring Further**:
The term \( p(p - 1) - q(q - 1) \) can be rewritten as:
\[ p^2 - p - (q^2 - q) = (p - q)(p + q - 1) \]
Thus, we have:
\[ 2a(p - q) + d(p - q)(p + q - 1) = 0 \]

7. **Factoring Out \( (p - q) \)**:
Factoring \( (p - q) \) out gives us:
\[ (p - q) \left( 2a + d(p + q - 1) \right) = 0 \]
This implies either \( p = q \) or \( 2a + d(p + q - 1) = 0 \).

8. **Finding the Sum of the First (p + q) Terms**:
We need to find \( S_{p + q} \):
\[ S_{p + q} = \frac{p + q}{2} \times (2a + (p + q - 1)d) \]
If \( 2a + d(p + q - 1) = 0 \), then:
\[ S_{p + q} = \frac{p + q}{2} \times 0 = 0 \]

### Conclusion:
Thus, the sum of the first (p + q) terms of the A.P. is:
\[ \boxed{0} \]