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 step by step, we will use the formula for the sum of the first n terms of an arithmetic progression (A.P.).

### Step 1: Write the formula for the sum of n terms of an A.P.
The sum of the first n terms (S_n) of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where:
- \(a\) is the first term,
- \(d\) is the common difference,
- \(n\) is the number of terms.

### Step 2: Set up the equation for the sum of 5 terms and 15 terms.
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:
\[ S_5 = \frac{5}{2} \times (2a + (5-1)d) = \frac{5}{2} \times (2a + 4d) \]
\[ S_{15} = \frac{15}{2} \times (2a + (15-1)d) = \frac{15}{2} \times (2a + 14d) \]

Setting these two equal:
\[ \frac{5}{2} \times (2a + 4d) = \frac{15}{2} \times (2a + 14d) \]

### Step 3: Simplify the equation.
We can eliminate \(\frac{1}{2}\) from both sides:
\[ 5(2a + 4d) = 15(2a + 14d) \]

Dividing both sides by 5:
\[ 2a + 4d = 3(2a + 14d) \]

Expanding the right side:
\[ 2a + 4d = 6a + 42d \]

### Step 4: Rearranging the equation.
Now, we will rearrange the equation to isolate terms involving \(a\) and \(d\):
\[ 2a + 4d - 6a - 42d = 0 \]
\[ -4a - 38d = 0 \]
\[ 4a = -38d \]
\[ a = -\frac{19}{2}d \]

### Step 5: Find the sum of the first 20 terms.
Now we need to find the sum of the first 20 terms \(S_{20}\):
\[ S_{20} = \frac{20}{2} \times (2a + (20-1)d) = 10 \times (2a + 19d) \]

Substituting \(a = -\frac{19}{2}d\) into the equation:
\[ S_{20} = 10 \times \left(2\left(-\frac{19}{2}d\right) + 19d\right) \]
\[ = 10 \times \left(-19d + 19d\right) \]
\[ = 10 \times 0 = 0 \]

### Conclusion
The sum of the first 20 terms of the A.P. is:
\[ \boxed{0} \]