To prove that \(\frac{1}{a}, b, \frac{1}{c}\) are in A.P. given that \(\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}\) are in A.P., we can follow these steps:
### Step 1: Understand the condition for A.P.
For three terms \(x, y, z\) to be in A.P., the condition is:
\[
y - x = z - y
\]
This can be rearranged to:
\[
2y = x + z
\]
### Step 2: Apply the A.P. condition to the given terms.
Let:
\[
x = \frac{a + b}{1 - ab}, \quad y = b, \quad z = \frac{b + c}{1 - bc}
\]
According to the A.P. condition:
\[
2b = \frac{a + b}{1 - ab} + \frac{b + c}{1 - bc}
\]
### Step 3: Clear the fractions.
Multiply both sides by \((1 - ab)(1 - bc)\) to eliminate the denominators:
\[
2b(1 - ab)(1 - bc) = (a + b)(1 - bc) + (b + c)(1 - ab)
\]
### Step 4: Expand both sides.
Expanding the left side:
\[
2b(1 - ab - bc + abc) = 2b - 2ab^2 - 2b^2c + 2ab^2c
\]
Expanding the right side:
\[
(a + b)(1 - bc) + (b + c)(1 - ab) = a + b - abc - b^2c + b + c - ab - abc
\]
Combining like terms:
\[
= (a + 2b + c) - (ab + 2abc + b^2c)
\]
### Step 5: Set the two sides equal.
Now we equate both expanded sides:
\[
2b - 2ab^2 - 2b^2c + 2ab^2c = a + 2b + c - (ab + 2abc + b^2c)
\]
### Step 6: Rearrange the equation.
Bringing all terms to one side gives:
\[
0 = a + c - ab - 2abc + 2b^2c - 2ab^2
\]
### Step 7: Factor and simplify.
Rearranging gives:
\[
2abc = a + c - ab + 2b^2c
\]
### Step 8: Divide by \(ac\).
Dividing the entire equation by \(ac\) yields:
\[
2b = \frac{1}{c} + \frac{1}{a}
\]
### Step 9: Conclude the proof.
This shows that:
\[
2b = \frac{1}{a} + \frac{1}{c}
\]
Thus, \(\frac{1}{a}, b, \frac{1}{c}\) are in A.P.
### Final Statement:
Therefore, we have proved that if \(\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}\) are in A.P., then \(\frac{1}{a}, b, \frac{1}{c}\) are also in A.P.
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