If a,b,c are in G.P then `(b-a)/(b-c)+(b+a)/(b+c)=`

Options

A. ab
B. ac
C. ` c^(2) - b^(2)`
D. 0

Answer

To solve the problem, we need to find the value of the expression \((b-a)/(b-c)+(b+a)/(b+c)\) given that \(a\), \(b\), and \(c\) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding G.P. Condition**:
Since \(a\), \(b\), and \(c\) are in G.P., we have the relationship:
\[ b^2 = ac \]
This means that \(b\) can be expressed as:
\[ b = \sqrt{ac} \]

2. **Substituting \(b\)**:
We substitute \(b\) into the expression \((b-a)/(b-c)+(b+a)/(b+c)\):
\[ \frac{\sqrt{ac} - a}{\sqrt{ac} - c} + \frac{\sqrt{ac} + a}{\sqrt{ac} + c} \]

3. **Simplifying the First Term**:
The first term is:
\[ \frac{\sqrt{ac} - a}{\sqrt{ac} - c} \]
We can rewrite it as:
\[ \frac{\sqrt{ac} - a}{\sqrt{ac} - c} = \frac{\sqrt{a}(\sqrt{c} - \sqrt{a})}{\sqrt{c}(\sqrt{a} - \sqrt{c})} \]

4. **Simplifying the Second Term**:
The second term is:
\[ \frac{\sqrt{ac} + a}{\sqrt{ac} + c} \]
Similarly, we can rewrite it as:
\[ \frac{\sqrt{a}(\sqrt{c} + \sqrt{a})}{\sqrt{c}(\sqrt{a} + \sqrt{c})} \]

5. **Combining the Two Terms**:
Now we have:
\[ \frac{\sqrt{a}(\sqrt{c} - \sqrt{a})}{\sqrt{c}(\sqrt{a} - \sqrt{c})} + \frac{\sqrt{a}(\sqrt{c} + \sqrt{a})}{\sqrt{c}(\sqrt{a} + \sqrt{c})} \]

6. **Finding a Common Denominator**:
The common denominator for both terms is \(\sqrt{c}(\sqrt{a} - \sqrt{c})(\sqrt{a} + \sqrt{c})\).

7. **Combining the Numerators**:
After combining the numerators, we simplify:
\[ \frac{\sqrt{a}((\sqrt{c} - \sqrt{a})(\sqrt{a} + \sqrt{c}) + (\sqrt{c} + \sqrt{a})(\sqrt{a} - \sqrt{c}))}{\sqrt{c}(\sqrt{a} - \sqrt{c})(\sqrt{a} + \sqrt{c})} \]

8. **Final Simplification**:
After simplification, we find that the numerator cancels out, leading to:
\[ 0 \]

### Final Result:
Thus, the value of the expression \((b-a)/(b-c)+(b+a)/(b+c)\) is:
\[ \boxed{0} \]