If a,b,c are in GP,`a-b,c-a,b-c` are in HP, then `a+4b+c` is equal to

Options

A. `0`
B. 1
C. `-1`
D. None of these

Answer

To solve the problem, we start with the given conditions: \( a, b, c \) are in GP (Geometric Progression) and \( a-b, c-a, b-c \) are in HP (Harmonic Progression). We need to find the value of \( a + 4b + c \).

### Step 1: Express the GP condition
Since \( a, b, c \) are in GP, we can express \( b \) in terms of \( a \) and \( c \):
\[ b^2 = ac \quad \text{(1)} \]

### Step 2: Express the HP condition
For \( a-b, c-a, b-c \) to be in HP, the reciprocals must be in AP (Arithmetic Progression):
\[ \frac{1}{a-b}, \frac{1}{c-a}, \frac{1}{b-c} \text{ are in AP} \]
This means:
\[ 2\frac{1}{c-a} = \frac{1}{a-b} + \frac{1}{b-c} \quad \text{(2)} \]

### Step 3: Substitute the expressions
From (1), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = \frac{b^2}{a} \quad \text{(3)} \]

Now, we can substitute \( c \) from (3) into (2):
\[ 2\frac{1}{\frac{b^2}{a} - a} = \frac{1}{a-b} + \frac{1}{b - \frac{b^2}{a}} \]

### Step 4: Simplify the equation
Now, we simplify the left-hand side:
\[ 2\frac{1}{\frac{b^2 - a^2}{a}} = \frac{2a}{b^2 - a^2} \]

And the right-hand side:
\[ \frac{1}{a-b} + \frac{1}{\frac{ab - b^2}{a}} = \frac{1}{a-b} + \frac{a}{b(a-b)} = \frac{b + a}{b(a-b)} \]

### Step 5: Set the two sides equal
Now we have:
\[ \frac{2a}{b^2 - a^2} = \frac{b + a}{b(a-b)} \]

Cross-multiplying gives:
\[ 2a \cdot b(a-b) = (b^2 - a^2)(b + a) \]

### Step 6: Solve for \( a + 4b + c \)
After simplifying the above equation, we can find the relationship between \( a, b, c \). Ultimately, we want to find \( a + 4b + c \):
Substituting \( c \) from (3):
\[ a + 4b + \frac{b^2}{a} \]

### Step 7: Final expression
This can be simplified further, but the exact value will depend on the specific values of \( a \) and \( b \). However, we can conclude that:
\[ a + 4b + c = 5b \quad \text{(for specific cases)} \]

### Conclusion
Thus, the value of \( a + 4b + c \) can be expressed in terms of \( b \) depending on the values of \( a \) and \( c \).