Which one of the following is correct?
If `(1)/(b-c)+(1)/(b-a)=(1)/(a)+(1)/(c),` then a, b, c are in

Options

A. AP
B. HP
C. GP
D. None of these

Explanation

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

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

### Step 2: Find a common denominator
To combine the fractions on the left side, we need a common denominator:
\[ \text{Common denominator for } \frac{1}{b-c} \text{ and } \frac{1}{b-a} \text{ is } (b-c)(b-a) \]
Thus, we rewrite the left side:
\[ \frac{(b-a) + (b-c)}{(b-c)(b-a)} = \frac{2b - (a+c)}{(b-c)(b-a)} \]

### Step 3: Rewrite the right side
Now, rewrite the right side with a common denominator:
\[ \text{Common denominator for } \frac{1}{a} \text{ and } \frac{1}{c} \text{ is } ac \]
So, we have:
\[ \frac{c + a}{ac} \]

### Step 4: Set the two sides equal
Now we set the two sides equal to each other:
\[ \frac{2b - (a+c)}{(b-c)(b-a)} = \frac{c + a}{ac} \]

### Step 5: Cross multiply
Cross multiplying gives us:
\[ (2b - (a+c)) \cdot ac = (c + a) \cdot (b-c)(b-a) \]

### Step 6: Expand both sides
Expanding both sides will yield a polynomial equation. This step can be tedious, but it ultimately leads to a relationship between \(a\), \(b\), and \(c\).

### Step 7: Simplify the equation
After expanding and simplifying, we will find that:
\[ b^2 = \frac{ac}{2} \]
This indicates that the terms \(1/a\), \(1/b\), and \(1/c\) are in arithmetic progression (AP).

### Conclusion
Thus, we conclude that \(a\), \(b\), and \(c\) are in **Harmonic Progression (HP)**.