If `H_(1)` and `H_(2)` are two harmonic means between two positive numbers a and b `(a != b)` , A and G are the arithmetic and geometric menas between a and b , then `(H_(2)+H_(1))/(H_(2)H_(1))` is

Options

A. `(2A)/(G)`
B. `A/(2G^(2))`
C. `A/G^(2)`
D. `(2A)/(G^(2))`

Explanation

To solve the problem, we need to find the value of \((H_2 + H_1) / (H_2 H_1)\) given that \(H_1\) and \(H_2\) are two harmonic means between two positive numbers \(a\) and \(b\) (where \(a \neq b\)), and \(A\) and \(G\) are the arithmetic and geometric means between \(a\) and \(b\).

### Step-by-Step Solution:

1. **Understanding Harmonic Means**:
The harmonic means \(H_1\) and \(H_2\) between \(a\) and \(b\) imply that the sequence \(a, H_1, H_2, b\) is in harmonic progression. This means that the reciprocals \(1/a, 1/H_1, 1/H_2, 1/b\) are in arithmetic progression (AP).

2. **Setting Up the AP**:
Let the common difference of this AP be \(d\). The first term is \(1/a\), so:
- \(1/H_1 = 1/a + d\)
- \(1/H_2 = 1/a + 2d\)
- \(1/b = 1/a + 3d\)

3. **Expressing \(d\)**:
From the equation for \(1/b\):
\[ 1/b = 1/a + 3d \implies 3d = 1/b - 1/a \implies d = \frac{1/b - 1/a}{3} \]

4. **Finding \(H_1\)**:
Substitute \(d\) back into the equation for \(1/H_1\):
\[ 1/H_1 = 1/a + d = 1/a + \frac{1/b - 1/a}{3} \]
Simplifying this gives:
\[ 1/H_1 = \frac{3/a + (1/b - 1/a)}{3} = \frac{(2/a + 1/b)}{3} \]
Thus, we have:
\[ H_1 = \frac{3ab}{2b + a} \]

5. **Finding \(H_2\)**:
Similarly, for \(H_2\):
\[ 1/H_2 = 1/a + 2d = 1/a + 2 \cdot \frac{1/b - 1/a}{3} \]
Simplifying gives:
\[ 1/H_2 = \frac{3/a + (2/b - 2/a)}{3} = \frac{(1/a + 2/b)}{3} \]
Thus, we have:
\[ H_2 = \frac{3ab}{a + 2b} \]

6. **Calculating \((H_2 + H_1) / (H_2 H_1)\)**:
Now we need to find:
\[ \frac{H_2 + H_1}{H_2 H_1} \]
First, calculate \(H_2 + H_1\):
\[ H_2 + H_1 = \frac{3ab}{a + 2b} + \frac{3ab}{2b + a} = 3ab \left(\frac{1}{a + 2b} + \frac{1}{2b + a}\right) \]
Finding a common denominator gives:
\[ H_2 + H_1 = 3ab \cdot \frac{(2b + a) + (a + 2b)}{(a + 2b)(2b + a)} = 3ab \cdot \frac{3(a + b)}{(a + 2b)(2b + a)} \]

7. **Finding \(H_2 H_1\)**:
\[ H_2 H_1 = \left(\frac{3ab}{a + 2b}\right) \left(\frac{3ab}{2b + a}\right) = \frac{9a^2b^2}{(a + 2b)(2b + a)} \]

8. **Final Calculation**:
Now substituting back:
\[ \frac{H_2 + H_1}{H_2 H_1} = \frac{3ab \cdot \frac{3(a + b)}{(a + 2b)(2b + a)}}{\frac{9a^2b^2}{(a + 2b)(2b + a)}} = \frac{3(a + b)}{ab} \]

### Conclusion:
Thus, the final result is:
\[ \frac{H_2 + H_1}{H_2 H_1} = \frac{3(a + b)}{ab} \]