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

Options

A. AP
B. HP
C. GP
D. Both (b) and (c )

Explanation

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

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

### Step 2: Combine the left-hand side
Find a common denominator for the left-hand side:
\[ \frac{(b-c) + (b-a)}{(b-a)(b-c)} = \frac{1}{a} + \frac{1}{c} \]
This simplifies to:
\[ \frac{2b - (a+c)}{(b-a)(b-c)} = \frac{1}{a} + \frac{1}{c} \]

### Step 3: Combine the right-hand side
The right-hand side can be combined as follows:
\[ \frac{c + a}{ac} \]

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

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

### Step 6: Expand both sides
Expand both sides:
- Left-hand side:
\[ 2abc - ac(a+c) \]
- Right-hand side:
\[ (a+c)(b^2 - (a+c)b + ac) \]

### Step 7: Rearrange and simplify
Rearranging and simplifying gives:
\[ 2abc - ac(a+c) = ab^2 - (a+c)ab + ac(a+c) \]

### Step 8: Collect like terms
After simplifying, we can collect like terms to find a relationship between \( a, b, \) and \( c \).

### Step 9: Determine the type of progression
From the derived equation, we can conclude that:
\[ 2ac = ab + bc \]
This indicates that the reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in arithmetic progression (AP). Therefore, \( a, b, c \) are in harmonic progression (HP).

### Final Conclusion
Thus, \( a, b, c \) are in HP.

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