Statement 1 If a and b be two positive numbers, where `agtb` and `4xxGM=5xxHM` for the numbers. Then, `a=4b`.
Statement 2 `(AM)(HM)=(GM)^(2)` is true for positive numbers.

Options

A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1
B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1
C. Statement 1 is true, Statement 2 is false
D. Statement 1 is false, Statement 2 is true

Explanation

To solve the problem step by step, we will analyze the two statements provided and derive the necessary conclusions.

### Step 1: Understanding the Means
We have two positive numbers \( a \) and \( b \) where \( a > b \). We need to find the relationships between the Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM).

1. **Arithmetic Mean (AM)** of \( a \) and \( b \):
\[ AM = \frac{a + b}{2} \]

2. **Geometric Mean (GM)** of \( a \) and \( b \):
\[ GM = \sqrt{ab} \]

3. **Harmonic Mean (HM)** of \( a \) and \( b \):
\[ HM = \frac{2ab}{a + b} \]

### Step 2: Using the Given Condition
The problem states that:
\[ 4 \times GM = 5 \times HM \]
Substituting the expressions for GM and HM:
\[ 4 \sqrt{ab} = 5 \left(\frac{2ab}{a + b}\right) \]
Simplifying this, we get:
\[ 4 \sqrt{ab} = \frac{10ab}{a + b} \]

### Step 3: Cross-Multiplying
Cross-multiplying gives:
\[ 4 \sqrt{ab} (a + b) = 10ab \]
Expanding this:
\[ 4a \sqrt{ab} + 4b \sqrt{ab} = 10ab \]

### Step 4: Rearranging the Equation
Rearranging the equation:
\[ 4a \sqrt{ab} + 4b \sqrt{ab} - 10ab = 0 \]

### Step 5: Factoring Out Common Terms
Factoring out \( 2\sqrt{ab} \):
\[ 2\sqrt{ab}(2a + 2b - 5\sqrt{ab}) = 0 \]
Since \( a \) and \( b \) are positive, \( \sqrt{ab} \neq 0 \). Therefore, we can set the other factor to zero:
\[ 2a + 2b - 5\sqrt{ab} = 0 \]

### Step 6: Solving for \( a \) in terms of \( b \)
Rearranging gives:
\[ 2a + 2b = 5\sqrt{ab} \]
Dividing by 2:
\[ a + b = \frac{5}{2} \sqrt{ab} \]
Squaring both sides:
\[ (a + b)^2 = \left(\frac{5}{2}\sqrt{ab}\right)^2 \]
Expanding both sides:
\[ a^2 + 2ab + b^2 = \frac{25}{4} ab \]
Rearranging gives:
\[ 4a^2 + 8ab + 4b^2 = 25ab \]
Simplifying:
\[ 4a^2 - 17ab + 4b^2 = 0 \]

### Step 7: Using the Quadratic Formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 4, B = -17b, C = 4b^2 \):
\[ D = (-17b)^2 - 4 \cdot 4 \cdot 4b^2 = 289b^2 - 64b^2 = 225b^2 \]
Thus:
\[ a = \frac{17b \pm 15b}{8} \]
This gives us two solutions:
1. \( a = 4b \)
2. \( a = \frac{1}{2}b \) (not valid since \( a > b \))

### Conclusion for Statement 1
Thus, we have proven that \( a = 4b \) is true.

### Step 8: Analyzing Statement 2
The second statement claims that:
\[ (AM)(HM) = (GM)^2 \]
This is true for positive numbers. We can verify this:
\[ AM \cdot HM = \left(\frac{a + b}{2}\right) \cdot \left(\frac{2ab}{a + b}\right) = ab \]
And:
\[ (GM)^2 = ab \]
Thus, the second statement is also true.

### Final Conclusion
- **Statement 1**: True
- **Statement 2**: True