The sum of first 5 terms and first 15 terms of an A.P. are equal. Find the sum of its first 20 terms.

Answer

To solve the problem, we need to find the sum of the first 20 terms of an arithmetic progression (A.P.) given that the sum of the first 5 terms and the first 15 terms are equal.

### Step-by-Step Solution:

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

2. **Setting Up the Equations**:
According to the problem, the sum of the first 5 terms is equal to the sum of the first 15 terms:
\[ S_5 = S_{15} \]
Using the formula for the sums:
\[ S_5 = \frac{5}{2} \times (2A + (5 - 1)D) = \frac{5}{2} \times (2A + 4D) \]
\[ S_{15} = \frac{15}{2} \times (2A + (15 - 1)D) = \frac{15}{2} \times (2A + 14D) \]

3. **Equating the Two Sums**:
Setting \( S_5 \) equal to \( S_{15} \):
\[ \frac{5}{2} \times (2A + 4D) = \frac{15}{2} \times (2A + 14D) \]

4. **Eliminating the Common Factor**:
We can multiply both sides by 2 to eliminate the fraction:
\[ 5(2A + 4D) = 15(2A + 14D) \]

5. **Expanding Both Sides**:
Expanding both sides gives:
\[ 10A + 20D = 30A + 210D \]

6. **Rearranging the Equation**:
Rearranging the equation to isolate terms involving \( A \) and \( D \):
\[ 10A - 30A = 210D - 20D \]
\[ -20A = 190D \]
Dividing by -10:
\[ 2A = -19D \quad \text{or} \quad A = -\frac{19}{2}D \]

7. **Finding the Sum of the First 20 Terms**:
Now we need to find the sum of the first 20 terms:
\[ S_{20} = \frac{20}{2} \times (2A + (20 - 1)D) = 10 \times (2A + 19D) \]
Substituting \( A = -\frac{19}{2}D \):
\[ S_{20} = 10 \times (2(-\frac{19}{2}D) + 19D) \]
Simplifying:
\[ = 10 \times (-19D + 19D) = 10 \times 0 = 0 \]

### Final Answer:
The sum of the first 20 terms of the A.P. is **0**.