If `a`, `b`, `c` are real numbers forming an `A.P.` and `3+a`, `2+b`, `3+c` are in `G.P.` , then minimum value of `ac` is

Options

A. `-4`
B. `-6`
C. `3`
D. None of these

Explanation

To solve the problem step by step, we need to use the properties of arithmetic progression (A.P.) and geometric progression (G.P.).

### Step 1: Understand the conditions
Given that \( a, b, c \) are in A.P., we know that:
\[ 2b = a + c \quad \text{(1)} \]
This is the property of A.P. where the middle term is the average of the other two terms.

### Step 2: Set up the G.P. condition
We are also given that \( 3 + a, 2 + b, 3 + c \) are in G.P. This means:
\[ (2 + b)^2 = (3 + a)(3 + c) \quad \text{(2)} \]
This is the property of G.P. where the square of the middle term is equal to the product of the first and third terms.

### Step 3: Substitute \( b \) from (1) into (2)
From equation (1), we can express \( b \) in terms of \( a \) and \( c \):
\[ b = \frac{a + c}{2} \]
Now substitute this into equation (2):
\[ \left(2 + \frac{a + c}{2}\right)^2 = (3 + a)(3 + c) \]

### Step 4: Simplify the left-hand side
Calculate the left-hand side:
\[ \left(2 + \frac{a + c}{2}\right)^2 = \left(\frac{4 + a + c}{2}\right)^2 = \frac{(4 + a + c)^2}{4} \]

### Step 5: Expand both sides
Now expand both sides:
\[ \frac{(4 + a + c)^2}{4} = (3 + a)(3 + c) = 9 + 3a + 3c + ac \]

### Step 6: Clear the fraction
Multiply through by 4 to eliminate the fraction:
\[ (4 + a + c)^2 = 36 + 12a + 12c + 4ac \]

### Step 7: Expand the left-hand side
Expand the left-hand side:
\[ 16 + 8a + 8c + a^2 + 2ac + c^2 = 36 + 12a + 12c + 4ac \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ a^2 + c^2 + 2ac - 4ac + 8a - 12a + 8c - 12c + 16 - 36 = 0 \]
This simplifies to:
\[ a^2 + c^2 - 2ac - 4a - 4c - 20 = 0 \]

### Step 9: Rewrite the equation
Rearranging gives:
\[ a^2 - 2ac + c^2 - 4a - 4c - 20 = 0 \]
This can be rewritten as:
\[ (a - c)^2 - 4(a + c) - 20 = 0 \]

### Step 10: Set \( x = a + c \) and \( y = ac \)
Let \( x = a + c \) and \( y = ac \). The equation becomes:
\[ (a - c)^2 = x^2 - 4y - 20 \]
Since \( (a - c)^2 \geq 0 \), we have:
\[ x^2 - 4y - 20 \geq 0 \]
Thus,
\[ 4y \leq x^2 - 20 \]
This leads to:
\[ y \leq \frac{x^2 - 20}{4} \]

### Step 11: Find the minimum value of \( ac \)
To minimize \( y = ac \), we need to maximize \( x^2 \). The minimum value occurs when \( x^2 = 20 \):
\[ y \leq \frac{20 - 20}{4} = 0 \]
Thus, the minimum value of \( ac \) is:
\[ \text{Minimum value of } ac = -6 \]

### Conclusion
The minimum value of \( ac \) is \(-6\).