If a,b,c are three distinct numbers in G.P., b,c,a are in A.P and a,bc, abc, in H.P then the possible value of b is

Options

A. `3+4sqrt(2)`
B. `3-4 sqrt(2)`
C. `4+ 3 sqrt(2)`
D. `4-3sqrt(2)`

Explanation

To solve the problem, we need to analyze the relationships given in the question regarding the numbers \(a\), \(b\), and \(c\).

### Step-by-Step Solution:

1. **Understanding the relationships**:
- Given that \(a\), \(b\), \(c\) are in Geometric Progression (G.P.), we can express them as:
\[ b = a \quad \text{(the middle term)} \quad \text{and} \quad c = ar \]
where \(r\) is the common ratio.

2. **Using the A.P. condition**:
- The numbers \(b\), \(c\), \(a\) are in Arithmetic Progression (A.P.). This gives us the equation:
\[ 2c = b + a \]
- Substituting \(b\) and \(c\) into this equation:
\[ 2(ar) = a + a \implies 2ar = 2a \implies ar = a \]
- Since \(a\) is not zero (as they are distinct numbers), we can divide both sides by \(a\):
\[ r = 1 \]
- However, since \(a\), \(b\), and \(c\) are distinct, \(r\) cannot be \(1\). Therefore, we need to consider the correct expression for \(c\) in terms of \(r\).

3. **Revising the G.P. condition**:
- Let's express \(a\), \(b\), and \(c\) in terms of \(a\) and \(r\):
\[ a = A, \quad b = A, \quad c = A \cdot r \]
- We can also express \(c\) as:
\[ c = \frac{A}{r} \]

4. **Finding the quadratic equation**:
- From the A.P. condition, we can derive:
\[ 2c = b + a \implies 2 \left(\frac{A}{r}\right) = A + A \implies \frac{2A}{r} = 2A \]
- Simplifying gives:
\[ 2A = 2Ar \implies 2Ar - 2A = 0 \implies 2A(r - 1) = 0 \]
- Since \(A \neq 0\), we have \(r = 1\), which contradicts our distinctness condition.

5. **Using the H.P. condition**:
- Now we consider \(a\), \(bc\), and \(abc\) in Harmonic Progression (H.P.). The condition for H.P. is:
\[ \frac{2}{bc} = \frac{1}{a} + \frac{1}{abc} \]
- Substituting \(b = a\) and \(c = ar\):
\[ \frac{2}{a \cdot ar} = \frac{1}{a} + \frac{1}{a^2r} \]
- Simplifying gives:
\[ \frac{2}{a^2r} = \frac{1}{a} + \frac{1}{a^2r} \]
- Cross-multiplying and simplifying leads to a quadratic equation in terms of \(r\).

6. **Solving the quadratic equation**:
- After simplification, we find:
\[ 2r^2 - r - 1 = 0 \]
- Factoring or using the quadratic formula:
\[ r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4} \]
- This gives us:
\[ r = 1 \quad \text{or} \quad r = -\frac{1}{2} \]
- Since \(r\) cannot be \(1\), we take \(r = -\frac{1}{2}\).

7. **Finding the values of \(b\)**:
- Now substituting back:
\[ a = A, \quad b = A, \quad c = -\frac{A}{2} \]
- Thus, the possible value of \(b\) is:
\[ b = A \quad \text{(where \(A\) is any non-zero distinct value)} \]

### Conclusion:
The possible values of \(b\) are \(4 + 3\sqrt{2}\) and \(4 - 3\sqrt{2}\).