The sum of 20 and 28 terms of an A.P. are equal. Find the sum of 48 terms of this A.P.

Explanation

To solve the problem, we need to find the sum of 48 terms of an arithmetic progression (A.P.) given that the sum of the first 20 terms is equal to the sum of the first 28 terms.

### Step-by-Step Solution:

1. **Understand 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:
\[ S_n = \frac{n}{2} \left( 2A + (n-1)D \right) \]
where \( A \) is the first term, \( D \) is the common difference, and \( n \) is the number of terms.

2. **Set up the equation based on the problem statement**:
We know that the sum of the first 20 terms is equal to the sum of the first 28 terms:
\[ S_{20} = S_{28} \]
Using the formula for the sums:
\[ \frac{20}{2} \left( 2A + (20-1)D \right) = \frac{28}{2} \left( 2A + (28-1)D \right) \]

3. **Simplify the equation**:
This simplifies to:
\[ 10 \left( 2A + 19D \right) = 14 \left( 2A + 27D \right) \]
Expanding both sides gives:
\[ 20A + 190D = 28A + 378D \]

4. **Rearranging the equation**:
Rearranging the equation leads to:
\[ 20A - 28A + 190D - 378D = 0 \]
Simplifying this results in:
\[ -8A - 188D = 0 \]
Dividing through by -4 gives:
\[ 2A + 47D = 0 \quad \text{(Equation 1)} \]

5. **Finding the sum of 48 terms**:
Now we need to find \( S_{48} \):
\[ S_{48} = \frac{48}{2} \left( 2A + (48-1)D \right) \]
This simplifies to:
\[ S_{48} = 24 \left( 2A + 47D \right) \]

6. **Substituting from Equation 1**:
From Equation 1, we know that \( 2A + 47D = 0 \). Therefore:
\[ S_{48} = 24 \times 0 = 0 \]

### Final Answer:
The sum of the first 48 terms of the A.P. is:
\[ \boxed{0} \]