Find three numbers in A.P. whose sum is 12 and product is 48.

Answer

To find three numbers in an Arithmetic Progression (A.P.) whose sum is 12 and product is 48, we can follow these steps:

### Step 1: Define the terms of the A.P.
Let the three numbers in A.P. be:
- First term: \( a - d \)
- Second term: \( a \)
- Third term: \( a + d \)

### Step 2: Set up the equation for the sum.
According to the problem, the sum of these three terms is given as:
\[ (a - d) + a + (a + d) = 12 \]
This simplifies to:
\[ 3a = 12 \]

### Step 3: Solve for \( a \).
Dividing both sides by 3, we find:
\[ a = 4 \]

### Step 4: Set up the equation for the product.
Next, we use the product condition:
\[ (a - d) \cdot a \cdot (a + d) = 48 \]
Substituting \( a = 4 \):
\[ (4 - d) \cdot 4 \cdot (4 + d) = 48 \]

### Step 5: Simplify the product equation.
This expands to:
\[ 4 \cdot (4^2 - d^2) = 48 \]
Thus:
\[ 4 \cdot (16 - d^2) = 48 \]
Dividing both sides by 4 gives:
\[ 16 - d^2 = 12 \]

### Step 6: Solve for \( d^2 \).
Rearranging gives:
\[ d^2 = 16 - 12 = 4 \]

### Step 7: Find the values of \( d \).
Taking the square root of both sides, we find:
\[ d = 2 \quad \text{or} \quad d = -2 \]

### Step 8: Determine the three numbers.
1. If \( d = 2 \):
- The numbers are \( 4 - 2 = 2 \), \( 4 \), and \( 4 + 2 = 6 \).
- So, one set of numbers is \( 2, 4, 6 \).

2. If \( d = -2 \):
- The numbers are \( 4 - (-2) = 6 \), \( 4 \), and \( 4 + (-2) = 2 \).
- So, another set of numbers is \( 6, 4, 2 \).

### Conclusion:
The three numbers in A.P. are \( 2, 4, 6 \) or \( 6, 4, 2 \).

---