`(1)/(b-a)+(1)/(b-c)=(1)/(a)+(1)/(c)` then a,b,c are in:

Options

A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer

To solve the equation \(\frac{1}{b-a} + \frac{1}{b-c} = \frac{1}{a} + \frac{1}{c}\) and determine the relationship among \(a\), \(b\), and \(c\), we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ \frac{1}{b-a} + \frac{1}{b-c} = \frac{1}{a} + \frac{1}{c} \]

### Step 2: Find a Common Denominator
For the left-hand side, the common denominator is \((b-a)(b-c)\):
\[ \frac{(b-c) + (b-a)}{(b-a)(b-c)} = \frac{1}{a} + \frac{1}{c} \]

### Step 3: Simplify the Left Side
Combine the terms in the numerator:
\[ \frac{(b-c) + (b-a)}{(b-a)(b-c)} = \frac{2b - (a+c)}{(b-a)(b-c)} \]

### Step 4: Simplify the Right Side
For the right-hand side, the common denominator is \(ac\):
\[ \frac{c + a}{ac} \]

### Step 5: Set the Two Sides Equal
Now we have:
\[ \frac{2b - (a+c)}{(b-a)(b-c)} = \frac{a+c}{ac} \]

### Step 6: Cross Multiply
Cross-multiply to eliminate the fractions:
\[ (2b - (a+c)) \cdot ac = (a+c) \cdot (b-a)(b-c) \]

### Step 7: Expand Both Sides
Expand both sides of the equation:
\[ 2abc - ac(a+c) = (a+c)(b^2 - (a+c)b + ac) \]

### Step 8: Rearrange the Equation
Rearranging the equation will help us isolate terms:
\[ 2abc - ac(a+c) = ab^2 + ac - a^2b - c^2b - acb \]

### Step 9: Factor and Simplify
After simplifying, we can factor out common terms and analyze the resulting equation.

### Step 10: Identify Relationships
From the simplifications, we can derive that \(b\) can be expressed in terms of \(a\) and \(c\). Specifically, we find:
\[ b = \frac{2ac}{a+c} \]

### Conclusion
The derived relationship indicates that \(a\), \(b\), and \(c\) are in Harmonic Progression (HP).

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