For `a`, `b`,`c in R-{0}`, let `(a+b)/(1-ab)`, `b`, `(b+c)/(1-bc)` are in `A.P.` If `alpha`, `beta` are the roots of the quadratic equation
`2acx^(2)+2abcx+(a+c)=0`, then the value of `(1+alpha)(1+beta)` is

Options

A. `0`
B. `1`
C. `-1`
D. `2`

Explanation

To solve the problem step by step, we will follow the logical progression based on the information provided in the question.

### Step 1: Understand the Arithmetic Progression Condition
We are given that the terms \(\frac{a+b}{1-ab}\), \(b\), and \(\frac{b+c}{1-bc}\) are in Arithmetic Progression (A.P.). For three numbers \(x\), \(y\), and \(z\) to be in A.P., the condition is:
\[ 2y = x + z \]
Applying this to our terms:
\[ 2b = \frac{a+b}{1-ab} + \frac{b+c}{1-bc} \]

### Step 2: Set Up the Equation
We can rearrange the equation:
\[ 2b(1-ab)(1-bc) = (a+b)(1-bc) + (b+c)(1-ab) \]

### Step 3: Expand Both Sides
Expanding both sides, we get:
\[ 2b(1 - ab - bc + abc) = (a + b - abc - b^2c) + (b + c - ab - b^2) \]

### Step 4: Simplify the Equation
After simplifying, we can collect like terms:
\[ 2b - 2ab^2 - 2bc + 2abc = 2b + c - ab - b^2 - abc \]
This leads us to:
\[ -a + c = 2abc \]
Thus, we have:
\[ a + c = 2abc \]

### Step 5: Use the Quadratic Equation
Now, we substitute \(a + c = 2abc\) into the quadratic equation:
\[ 2acx^2 + 2abcx + (a + c) = 0 \]
Substituting \(a + c\):
\[ 2acx^2 + 2abcx + 2abc = 0 \]

### Step 6: Factor Out Common Terms
Factoring out \(2ac\):
\[ 2ac(x^2 + \frac{abc}{ac}x + 1) = 0 \]
This gives us:
\[ x^2 + \frac{b}{a}x + 1 = 0 \]

### Step 7: Find the Roots
Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation. By Vieta's formulas:
\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = 1 \]

### Step 8: Calculate \( (1 + \alpha)(1 + \beta) \)
We need to find:
\[ (1 + \alpha)(1 + \beta) = 1 + \alpha + \beta + \alpha \beta \]
Substituting the values from Vieta's:
\[ = 1 - \frac{b}{a} + 1 = 2 - \frac{b}{a} \]

### Step 9: Conclusion
Since we know that \(\alpha \beta = 1\), we can conclude:
\[ (1 + \alpha)(1 + \beta) = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]