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 step by step, 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 1: Write 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. 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.

### Step 2: Calculate the sum of the first 8 terms \( S_8 \)
Using the formula:
\[ S_8 = \frac{8}{2} \times (2A + (8-1)D) = 4 \times (2A + 7D) = 8A + 28D \]

### Step 3: Calculate the sum of the first 24 terms \( S_{24} \)
Using the formula:
\[ S_{24} = \frac{24}{2} \times (2A + (24-1)D) = 12 \times (2A + 23D) = 24A + 276D \]

### Step 4: Set the sums equal to each other
According to the problem, the sum of the first 8 terms is equal to the sum of the first 24 terms:
\[ S_8 = S_{24} \]
Thus, we have:
\[ 8A + 28D = 24A + 276D \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 8A - 24A + 28D - 276D = 0 \]
This simplifies to:
\[ -16A - 248D = 0 \]
Dividing through by -16 gives:
\[ A + 15.5D = 0 \quad \text{(1)} \]

### Step 6: Calculate the sum of the first 32 terms \( S_{32} \)
Using the formula for \( S_{32} \):
\[ S_{32} = \frac{32}{2} \times (2A + (32-1)D) = 16 \times (2A + 31D) = 32A + 496D \]

### Step 7: Substitute \( A \) from equation (1)
From equation (1), we can express \( A \) in terms of \( D \):
\[ A = -15.5D \]
Substituting this into the equation for \( S_{32} \):
\[ S_{32} = 32(-15.5D) + 496D = -496D + 496D = 0 \]

### Conclusion
Thus, the sum of the first 32 terms of the A.P. is:
\[ \boxed{0} \]