The sum of the first three terms of an A.P. is 36 while their product is 1620. Find the A.P.

Answer

To find the arithmetic progression (A.P.) where the sum of the first three terms is 36 and their product is 1620, we can follow these steps:

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

### Step 2: Set up the equation for the sum of the terms
According to the problem, the sum of the first three terms is 36:
\[ (a - d) + a + (a + d) = 36 \]
This simplifies to:
\[ 3a = 36 \]
Thus, we can find \( a \):
\[ a = \frac{36}{3} = 12 \]

### Step 3: Set up the equation for the product of the terms
Next, we know that the product of the first three terms is 1620:
\[ (a - d) \cdot a \cdot (a + d) = 1620 \]
Substituting \( a = 12 \) into the equation:
\[ (12 - d) \cdot 12 \cdot (12 + d) = 1620 \]

### Step 4: Simplify the product equation
Expanding the product:
\[ 12 \cdot (12^2 - d^2) = 1620 \]
This simplifies to:
\[ 12 \cdot (144 - d^2) = 1620 \]
Dividing both sides by 12:
\[ 144 - d^2 = \frac{1620}{12} \]
Calculating the right side:
\[ 144 - d^2 = 135 \]

### Step 5: Solve for \( d^2 \)
Rearranging gives:
\[ d^2 = 144 - 135 = 9 \]
Taking the square root:
\[ d = 3 \quad \text{or} \quad d = -3 \]

### Step 6: Find the A.P. for both values of \( d \)
1. **If \( d = 3 \)**:
- First term: \( a - d = 12 - 3 = 9 \)
- Second term: \( a = 12 \)
- Third term: \( a + d = 12 + 3 = 15 \)
- A.P.: \( 9, 12, 15, \ldots \)

2. **If \( d = -3 \)**:
- First term: \( a - d = 12 - (-3) = 15 \)
- Second term: \( a = 12 \)
- Third term: \( a + d = 12 + (-3) = 9 \)
- A.P.: \( 15, 12, 9, \ldots \)

### Final Answer
The two possible arithmetic progressions are:
1. \( 9, 12, 15, \ldots \)
2. \( 15, 12, 9, \ldots \)