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 step by step, we will use the formula for the sum of the first \( n \) terms of an arithmetic progression (A.P.) and the information given in the question.

### Step-by-Step Solution:

1. **Understanding the Sum of Terms in A.P.**:
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
where \( A \) is the first term, \( D \) is the common difference, and \( n \) is the number of terms.

2. **Setting Up the Given Condition**:
We are given that the sum of the first \( p \) terms is equal to the sum of the first \( q \) terms:
\[ S_p = S_q \]
Therefore, we can write:
\[ \frac{p}{2} \left(2A + (p-1)D\right) = \frac{q}{2} \left(2A + (q-1)D\right) \]

3. **Removing 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:
\[ 2Ap + p(p-1)D = 2Aq + q(q-1)D \]

5. **Rearranging the Equation**:
Bringing all terms to one side:
\[ 2Ap - 2Aq + p(p-1)D - q(q-1)D = 0 \]
This can be factored as:
\[ 2A(p - q) + D(p^2 - q^2 - p + q) = 0 \]

6. **Factoring Further**:
Recognizing that \( p^2 - q^2 \) can be factored as \( (p - q)(p + q) \):
\[ 2A(p - q) + D((p - q)(p + q)) = 0 \]
Factoring out \( (p - q) \):
\[ (p - q)(2A + D(p + q)) = 0 \]

7. **Analyzing the Factors**:
Since \( p \) and \( q \) are not equal (as stated in the problem), we can conclude:
\[ 2A + D(p + q) = 0 \]
Rearranging gives:
\[ 2A = -D(p + q) \]

8. **Finding the Sum of the First \( p + q \) Terms**:
Now, we need to find the sum of the first \( p + q \) terms:
\[ S_{p+q} = \frac{p+q}{2} \left(2A + (p + q - 1)D\right) \]
Substituting \( 2A = -D(p + q) \):
\[ S_{p+q} = \frac{p+q}{2} \left(-D(p + q) + (p + q - 1)D\right) \]
Simplifying inside the parentheses:
\[ S_{p+q} = \frac{p+q}{2} \left(-D(p + q) + D(p + q - 1)\right) \]
\[ = \frac{p+q}{2} \left(-D(p + q) + D(p + q) - D\right) \]
\[ = \frac{p+q}{2} \left(-D\right) \]
Thus:
\[ S_{p+q} = -\frac{(p + q)D}{2} \]

9. **Conclusion**:
Therefore, the sum of the first \( (p + q) \) terms is:
\[ S_{p+q} = 0 \]