The A.M. of two numbers exceeds the G.M. by 2 and the G.M. exceeds the H.M. by `8/5,` find the numbers.

Answer

To solve the problem, we need to find two numbers \( a \) and \( b \) given the conditions regarding their Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.).

### Step 1: Define the Means
1. **Arithmetic Mean (A.M.)**:
\[ A.M. = \frac{a + b}{2} \]

2. **Geometric Mean (G.M.)**:
\[ G.M. = \sqrt{ab} \]

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

### Step 2: Set Up the Equations
From the problem statement, we have:
1. The A.M. exceeds the G.M. by 2:
\[ \frac{a + b}{2} = G.M. + 2 \]
Substituting for G.M.:
\[ \frac{a + b}{2} = \sqrt{ab} + 2 \]

2. The G.M. exceeds the H.M. by \( \frac{8}{5} \):
\[ \sqrt{ab} = H.M. + \frac{8}{5} \]
Substituting for H.M.:
\[ \sqrt{ab} = \frac{2ab}{a + b} + \frac{8}{5} \]

### Step 3: Solve the First Equation
From the first equation:
\[ \frac{a + b}{2} - 2 = \sqrt{ab} \]
Multiplying through by 2:
\[ a + b - 4 = 2\sqrt{ab} \]
Rearranging gives:
\[ a + b = 2\sqrt{ab} + 4 \quad \text{(1)} \]

### Step 4: Solve the Second Equation
From the second equation:
\[ \sqrt{ab} - \frac{8}{5} = \frac{2ab}{a + b} \]
Multiplying through by \( a + b \):
\[ (a + b)\sqrt{ab} - \frac{8}{5}(a + b) = 2ab \]
Substituting \( a + b \) from equation (1):
\[ (2\sqrt{ab} + 4)\sqrt{ab} - \frac{8}{5}(2\sqrt{ab} + 4) = 2ab \]
Expanding this gives:
\[ 2ab + 4\sqrt{ab} - \frac{16\sqrt{ab}}{5} - \frac{32}{5} = 2ab \]
Simplifying:
\[ 4\sqrt{ab} - \frac{16\sqrt{ab}}{5} - \frac{32}{5} = 0 \]
Multiplying through by 5 to eliminate the fraction:
\[ 20\sqrt{ab} - 16\sqrt{ab} - 32 = 0 \]
This simplifies to:
\[ 4\sqrt{ab} = 32 \]
Thus:
\[ \sqrt{ab} = 8 \quad \Rightarrow \quad ab = 64 \quad \text{(2)} \]

### Step 5: Substitute Back to Find \( a \) and \( b \)
Now we have two equations:
1. \( a + b = 20 \) (from substituting \( G.M. \) back into equation (1))
2. \( ab = 64 \)

We can use these to form a quadratic equation:
\[ x^2 - (a + b)x + ab = 0 \]
Substituting the values:
\[ x^2 - 20x + 64 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here \( a = 1, b = -20, c = 64 \):
\[ x = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 20^2 - 256 = 400 - 256 = 144 \]
Thus:
\[ x = \frac{20 \pm 12}{2} \]
Calculating the roots:
1. \( x = \frac{32}{2} = 16 \)
2. \( x = \frac{8}{2} = 4 \)

### Final Result
The two numbers are \( a = 16 \) and \( b = 4 \).