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

Explanation

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 is equal to the sum of the first 15 terms.

### Step-by-Step Solution:

1. **Understand the formula for the sum of the first n terms of an A.P.**:
The formula for the sum of the first n terms (S_n) 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,
- \( n \) is the number of terms.

2. **Set up the equations for S_5 and S_15**:
Since we know that \( S_5 = S_{15} \):
\[ S_5 = \frac{5}{2} \times (2A + (5-1)D) = \frac{5}{2} \times (2A + 4D) \]
\[ S_{15} = \frac{15}{2} \times (2A + (15-1)D) = \frac{15}{2} \times (2A + 14D) \]

3. **Set the two sums equal**:
\[ \frac{5}{2} \times (2A + 4D) = \frac{15}{2} \times (2A + 14D) \]

4. **Eliminate the common factor**:
We can multiply both sides by 2 to eliminate the fraction:
\[ 5 \times (2A + 4D) = 15 \times (2A + 14D) \]

5. **Distribute**:
\[ 10A + 20D = 30A + 210D \]

6. **Rearrange the equation**:
Move all terms involving \( A \) to one side and terms involving \( D \) to the other:
\[ 10A - 30A = 210D - 20D \]
\[ -20A = 190D \]
\[ A = -\frac{19}{2}D \]

7. **Find S_20**:
Now we can find the sum of the first 20 terms \( S_{20} \):
\[ S_{20} = \frac{20}{2} \times (2A + (20-1)D) = 10 \times (2A + 19D) \]
Substitute \( A = -\frac{19}{2}D \):
\[ S_{20} = 10 \times \left(2 \left(-\frac{19}{2}D\right) + 19D\right) \]
\[ = 10 \times \left(-19D + 19D\right) = 10 \times 0 = 0 \]

### Final Answer:
The sum of the first 20 terms of the A.P. is \( S_{20} = 0 \).