If `A_(1),A_(2)` are between two numbers, then `(A_(1)+A_(2))/(H_(1)+H_(2))` is equal to

Options

A. `(H_(1)H_(2))/(G_(1)G_(2))`
B. `(G_(1)G_(2))/(H_(1)H_(2))`
C. `(H_(1)H_(2))/(A_(1)A_(2))`
D. `(G_(1)G_(2))/(A_(1)A_(2))`

Answer

To solve the problem, we need to find the value of \(\frac{A_1 + A_2}{H_1 + H_2}\) given that \(A_1\) and \(A_2\) are arithmetic means between two numbers \(a\) and \(b\), and \(H_1\) and \(H_2\) are harmonic means between the same two numbers.

### Step-by-Step Solution:

1. **Understanding the Means**:
- Since \(A_1\) and \(A_2\) are arithmetic means between \(a\) and \(b\), we can express them as:
\[ A_1 = a + d \quad \text{and} \quad A_2 = a + 2d \]
where \(d\) is the common difference.

2. **Sum of Arithmetic Means**:
- The sum of \(A_1\) and \(A_2\) can be calculated as:
\[ A_1 + A_2 = (a + d) + (a + 2d) = 2a + 3d \]

3. **Finding \(H_1\) and \(H_2\)**:
- The harmonic means \(H_1\) and \(H_2\) between \(a\) and \(b\) can be expressed using the formula for harmonic means:
\[ H_1 = \frac{2ab}{a + b} \quad \text{and} \quad H_2 = \frac{2ab}{a + b} \]
(Note: In this case, both harmonic means are the same since they are between the same two numbers.)

4. **Sum of Harmonic Means**:
- Therefore, we have:
\[ H_1 + H_2 = \frac{2ab}{a + b} + \frac{2ab}{a + b} = \frac{4ab}{a + b} \]

5. **Substituting into the Expression**:
- Now, we substitute \(A_1 + A_2\) and \(H_1 + H_2\) into our original expression:
\[ \frac{A_1 + A_2}{H_1 + H_2} = \frac{2a + 3d}{\frac{4ab}{a + b}} \]

6. **Simplifying the Expression**:
- To simplify, we can multiply by the reciprocal:
\[ = \frac{(2a + 3d)(a + b)}{4ab} \]

7. **Final Result**:
- This gives us the required expression for \(\frac{A_1 + A_2}{H_1 + H_2}\).

### Conclusion:
The value of \(\frac{A_1 + A_2}{H_1 + H_2}\) is given by:
\[ \frac{(2a + 3d)(a + b)}{4ab} \]