The sum of three numbers in A.P. is 27, and their product is 504, find them.

Answer

To solve the problem, we need to find three numbers in Arithmetic Progression (A.P.) whose sum is 27 and whose product is 504. Let's denote the three numbers as \( a - d \), \( a \), and \( a + d \), where \( a \) is the middle term and \( d \) is the common difference.

### Step-by-step Solution:

1. **Set up the equations based on the given conditions:**
- The sum of the three numbers is given by:
\[ (a - d) + a + (a + d) = 27 \]
- Simplifying this, we get:
\[ 3a = 27 \]
- Therefore, solving for \( a \):
\[ a = \frac{27}{3} = 9 \]

2. **Set up the product equation:**
- The product of the three numbers is given by:
\[ (a - d) \cdot a \cdot (a + d) = 504 \]
- Substituting \( a = 9 \):
\[ (9 - d) \cdot 9 \cdot (9 + d) = 504 \]

3. **Simplify the product equation:**
- This can be rewritten as:
\[ 9 \cdot (9 - d) \cdot (9 + d) = 504 \]
- Using the difference of squares:
\[ 9 \cdot (9^2 - d^2) = 504 \]
- Simplifying further:
\[ 9 \cdot (81 - d^2) = 504 \]
- Dividing both sides by 9:
\[ 81 - d^2 = \frac{504}{9} = 56 \]

4. **Solve for \( d^2 \):**
- Rearranging gives:
\[ d^2 = 81 - 56 = 25 \]
- Taking the square root:
\[ d = 5 \quad \text{or} \quad d = -5 \]

5. **Find the three numbers:**
- **Case 1:** If \( d = 5 \):
- First number: \( a - d = 9 - 5 = 4 \)
- Second number: \( a = 9 \)
- Third number: \( a + d = 9 + 5 = 14 \)
- The three numbers are \( 4, 9, 14 \).

- **Case 2:** If \( d = -5 \):
- First number: \( a - d = 9 - (-5) = 14 \)
- Second number: \( a = 9 \)
- Third number: \( a + d = 9 + (-5) = 4 \)
- The three numbers are \( 14, 9, 4 \).

6. **Conclusion:**
- The three numbers in A.P. are \( 4, 9, 14 \) (the order does not matter).