If sum of first 20 terms of an A.P. is equal to sum of first 30 terms of the A.P. then sum of the first 50 terms of the A.P. is

Options

A. `-1`
B. 0
C. 10
D. 25

Answer

To solve the problem, we need to find the sum of the first 50 terms of an arithmetic progression (A.P.) given that the sum of the first 20 terms is equal to the sum of the first 30 terms.

### Step-by-step Solution:

1. **Understanding the Formula for the Sum of the First n Terms of an A.P.**:
The sum of the first n terms (S_n) of an A.P. can be expressed as:
\[ S_n = \frac{n}{2} \times [2a + (n - 1)d] \]
where:
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

2. **Setting Up the Equations**:
Given that the sum of the first 20 terms is equal to the sum of the first 30 terms, we can write:
\[ S_{20} = S_{30} \]
Using the formula, we can express this as:
\[ \frac{20}{2} \times [2a + (20 - 1)d] = \frac{30}{2} \times [2a + (30 - 1)d] \]
Simplifying this gives:
\[ 10 \times [2a + 19d] = 15 \times [2a + 29d] \]

3. **Expanding and Rearranging**:
Expanding both sides:
\[ 20a + 190d = 30a + 435d \]
Rearranging gives:
\[ 20a - 30a + 190d - 435d = 0 \]
Simplifying further:
\[ -10a - 245d = 0 \]
Dividing through by -5:
\[ 2a + 49d = 0 \quad \text{(Equation 1)} \]

4. **Finding the Sum of the First 50 Terms**:
Now, we need to find \( S_{50} \):
\[ S_{50} = \frac{50}{2} \times [2a + (50 - 1)d] \]
This simplifies to:
\[ S_{50} = 25 \times [2a + 49d] \]
From Equation 1, we know that \( 2a + 49d = 0 \). Therefore:
\[ S_{50} = 25 \times 0 = 0 \]

### Conclusion:
The sum of the first 50 terms of the A.P. is:
\[ \boxed{0} \]