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

Explanation

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

### Step 1: Define the three numbers in A.P.
Let the three numbers be \( A - D \), \( A \), and \( A + D \), where \( A \) is the middle term and \( D \) is the common difference.

### Step 2: Set up the equation for the sum.
The sum of these three numbers is given as:
\[ (A - D) + A + (A + D) = 12 \]
Simplifying this, we get:
\[ 3A = 12 \]
Thus,
\[ A = 4 \]

### Step 3: Set up the equation for the product.
The product of the three numbers is given as:
\[ (A - D) \cdot A \cdot (A + D) = 48 \]
Substituting \( A = 4 \):
\[ (4 - D) \cdot 4 \cdot (4 + D) = 48 \]
This simplifies to:
\[ 4(4 - D)(4 + D) = 48 \]
Dividing both sides by 4:
\[ (4 - D)(4 + D) = 12 \]
Using the difference of squares:
\[ 16 - D^2 = 12 \]
Thus,
\[ D^2 = 4 \]
So,
\[ D = 2 \quad \text{or} \quad D = -2 \]

### Step 4: Find the three numbers.
1. If \( D = 2 \):
- The numbers are:
\[ A - D = 4 - 2 = 2, \quad A = 4, \quad A + D = 4 + 2 = 6 \]
So, the numbers are \( 2, 4, 6 \).

2. If \( D = -2 \):
- The numbers are:
\[ A - D = 4 - (-2) = 4 + 2 = 6, \quad A = 4, \quad A + D = 4 + (-2) = 4 - 2 = 2 \]
So, the numbers are also \( 6, 4, 2 \).

### Conclusion:
The three numbers in A.P. whose sum is 12 and product is 48 are \( 2, 4, 6 \).

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