Two A.M.'s `A_(1) and A_(2)`, two G.M.'s `G_(1) and G_(2)` and two H.M's `H_(1) and H_(2)` are inserted between any two numbers, then `H_(1)^(-1) + H_(2)^(-1)` equals

Options

A. `A_(1)^(-1) + A_(2)^(-1)`
B. `G_(1)^(-1) + G_(2)^(-1)`
C. `(G_(1)G_(2))/(A_(1) + A_(2))`
D. `(A_(1) + A_(2))/(G_(1)G_(2))`

Answer

To solve the problem, we need to find the value of \( H_1^{-1} + H_2^{-1} \) given that two A.M.s, two G.M.s, and two H.M.s are inserted between two numbers \( a \) and \( b \).

### Step-by-Step Solution:

1. **Define the Terms**:
Let the two numbers be \( a \) and \( b \). The two A.M.s inserted between them are \( A_1 \) and \( A_2 \), the two G.M.s are \( G_1 \) and \( G_2 \), and the two H.M.s are \( H_1 \) and \( H_2 \).

2. **Using A.M. (Arithmetic Mean)**:
Since \( A_1 \) and \( A_2 \) are the A.M.s between \( a \) and \( b \):
\[ \frac{a + b}{2} = \frac{A_1 + A_2}{2} \]
Thus, we have:
\[ A_1 + A_2 = a + b \]
(Equation 1)

3. **Using G.M. (Geometric Mean)**:
Since \( G_1 \) and \( G_2 \) are the G.M.s between \( a \) and \( b \):
\[ \sqrt{ab} = \sqrt{G_1 \cdot G_2} \]
Squaring both sides gives:
\[ ab = G_1 \cdot G_2 \]
(Equation 2)

4. **Using H.M. (Harmonic Mean)**:
Since \( H_1 \) and \( H_2 \) are the H.M.s between \( a \) and \( b \):
\[ \frac{1}{H_1} + \frac{1}{H_2} = \frac{2}{\frac{1}{\frac{1}{a} + \frac{1}{b}}} \]
This can be rewritten as:
\[ \frac{1}{H_1} + \frac{1}{H_2} = \frac{2ab}{a + b} \]

5. **Relating \( H_1^{-1} + H_2^{-1} \) to \( A_1, A_2, G_1, G_2 \)**:
From the previous steps, we know:
\[ H_1^{-1} + H_2^{-1} = \frac{2ab}{a + b} \]
Now substituting \( a + b \) from Equation 1 and \( ab \) from Equation 2:
\[ H_1^{-1} + H_2^{-1} = \frac{2G_1G_2}{A_1 + A_2} \]

6. **Final Expression**:
Therefore, we can conclude:
\[ H_1^{-1} + H_2^{-1} = \frac{A_1 + A_2}{G_1G_2} \]

### Conclusion:
The final answer is:
\[ H_1^{-1} + H_2^{-1} = \frac{A_1 + A_2}{G_1G_2} \]