The sum of first 8 terms and first 24 terms of an A.P. are equal. Find the sum of its 32 terms.

Answer

To solve the problem, we need to find the sum of the first 32 terms of an arithmetic progression (A.P.) given that the sum of the first 8 terms is equal to the sum of the first 24 terms.

### Step-by-Step Solution:

1. **Understanding 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 and \( d \) is the common difference.

2. **Setting Up the Equations**:
From the problem, we know:
\[ S_8 = S_{24} \]
Therefore, we can write:
\[ S_8 = \frac{8}{2} \times (2a + (8 - 1)d) = 4 \times (2a + 7d) = 8a + 28d \]
and
\[ S_{24} = \frac{24}{2} \times (2a + (24 - 1)d) = 12 \times (2a + 23d) = 24a + 276d \]

3. **Setting the Two Sums Equal**:
Since \( S_8 = S_{24} \):
\[ 8a + 28d = 24a + 276d \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ 8a - 24a + 28d - 276d = 0 \]
Simplifying this, we have:
\[ -16a - 248d = 0 \]
or
\[ 16a = -248d \]
Dividing both sides by 16:
\[ a = -\frac{248}{16}d = -\frac{31}{2}d \]

5. **Finding the Sum of the First 32 Terms**:
Now we can find \( S_{32} \):
\[ S_{32} = \frac{32}{2} \times (2a + (32 - 1)d) = 16 \times (2a + 31d) \]
Substituting \( a = -\frac{31}{2}d \):
\[ S_{32} = 16 \times \left(2 \left(-\frac{31}{2}d\right) + 31d\right) \]
Simplifying inside the parentheses:
\[ S_{32} = 16 \times \left(-31d + 31d\right) = 16 \times 0 = 0 \]

### Final Answer:
The sum of the first 32 terms of the A.P. is \( 0 \).