The A.M., G.M. and H.M. between two positive numbers a and b are equal, then

Options

A. a = b
B. ab = 1
C. `a gt b`
D. `a lt b`

Answer

To solve the problem, we need to show that if the Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.) of two positive numbers \( a \) and \( b \) are equal, then \( a \) must be equal to \( b \).

### Step-by-Step Solution:

1. **Define the Means**:
- The Arithmetic Mean (A.M.) of \( a \) and \( b \) is given by:
\[ A.M. = \frac{a + b}{2} \]
- The Geometric Mean (G.M.) of \( a \) and \( b \) is given by:
\[ G.M. = \sqrt{ab} \]
- The Harmonic Mean (H.M.) of \( a \) and \( b \) is given by:
\[ H.M. = \frac{2ab}{a + b} \]

2. **Set the Means Equal**:
Since we are given that A.M., G.M., and H.M. are equal, we can write:
\[ A.M. = G.M. = H.M. \]
Therefore, we have:
\[ \frac{a + b}{2} = \sqrt{ab} = \frac{2ab}{a + b} \]

3. **Equate A.M. and G.M.**:
First, we equate A.M. and G.M.:
\[ \frac{a + b}{2} = \sqrt{ab} \]
Squaring both sides gives:
\[ \left(\frac{a + b}{2}\right)^2 = ab \]
Expanding the left side:
\[ \frac{(a + b)^2}{4} = ab \]
Multiplying both sides by 4:
\[ (a + b)^2 = 4ab \]
Expanding the left side:
\[ a^2 + 2ab + b^2 = 4ab \]
Rearranging gives:
\[ a^2 + b^2 - 2ab = 0 \]
This can be factored as:
\[ (a - b)^2 = 0 \]
Therefore, we conclude:
\[ a - b = 0 \implies a = b \]

4. **Equate A.M. and H.M.**:
Now we can also check A.M. and H.M. for completeness:
\[ \frac{a + b}{2} = \frac{2ab}{a + b} \]
Cross-multiplying gives:
\[ (a + b)^2 = 4ab \]
This is the same equation we derived earlier, confirming that \( a = b \).

### Conclusion:
Thus, if the A.M., G.M., and H.M. of two positive numbers \( a \) and \( b \) are equal, then \( a \) must be equal to \( b \).