If `(1)/(a)+(1)/(c )=(1)/(2b-a)+(1)/(2b-c)`, then

Options

A. `a,b,c` are in `A.P.`
B. `a,(b)/(2),c` are in `A.P.`
C. `a,(b)/(2),c` are in `H.P.`
D. `a,2b,c` are in `H.P.`

Explanation

To solve the equation \(\frac{1}{a} + \frac{1}{c} = \frac{1}{2b-a} + \frac{1}{2b-c}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{1}{a} + \frac{1}{c} - \frac{1}{2b-a} - \frac{1}{2b-c} = 0 \]

### Step 2: Combine the fractions
To combine the fractions, we need a common denominator. The common denominator for the left side is \(ac\) and for the right side is \((2b-a)(2b-c)\). Thus, we can rewrite the equation as:
\[ \frac{c + a}{ac} - \left(\frac{(2b-c) + (2b-a)}{(2b-a)(2b-c)}\right) = 0 \]

### Step 3: Simplify the right side
The right side simplifies to:
\[ \frac{(2b-c) + (2b-a)}{(2b-a)(2b-c)} = \frac{4b - (a+c)}{(2b-a)(2b-c)} \]

### Step 4: Set the equation to zero
Now we have:
\[ \frac{c + a}{ac} - \frac{4b - (a+c)}{(2b-a)(2b-c)} = 0 \]
This implies:
\[ \frac{c + a}{ac} = \frac{4b - (a+c)}{(2b-a)(2b-c)} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ (c + a)(2b - a)(2b - c) = ac(4b - (a + c)) \]

### Step 6: Expand both sides
Expanding both sides leads to:
\[ (c + a)(4b - (a + c)) = ac(4b - (a + c)) \]

### Step 7: Factor out common terms
Factoring out common terms gives:
\[ (c + a)(2b - a)(2b - c) = 0 \]

### Step 8: Analyze the factors
This leads to two scenarios:
1. \(c + a = 0\) which is not possible since \(a\) and \(c\) are positive.
2. \(2b - a = 0\) or \(2b - c = 0\).

### Step 9: Solve for \(b\)
From \(2b - a = 0\), we have:
\[ b = \frac{a}{2} \]
From \(2b - c = 0\), we have:
\[ b = \frac{c}{2} \]

### Step 10: Conclusion
Thus, we find that \(a + c = 2b\), which means \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP).

### Final Result
The relation \(a + c = 2b\) confirms that \(a\), \(b\), and \(c\) are in AP.

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