The arithmetic mean of two numbers exceeds their geometric mean by 2 and the geometric mean exceeds their harmonic mean by 1.6. What are the two numbers ?

Options

A. 16,4
B. 81,9
C. 256, 16
D. 625,25

Explanation

To solve the problem, we need to find two numbers based on the relationships between their arithmetic mean (A), geometric mean (G), and harmonic mean (H). Let's denote the two numbers as \( a \) and \( b \).

### Step 1: Set up the equations based on the problem statement

1. The arithmetic mean exceeds the geometric mean by 2:
\[ A = G + 2 \]

2. The geometric mean exceeds the harmonic mean by 1.6:
\[ G = H + 1.6 \]

### Step 2: Express A in terms of H

From the second equation, we can express \( G \) in terms of \( H \):
\[ G = H + 1.6 \]

Now substituting this into the first equation:
\[ A = (H + 1.6) + 2 = H + 3.6 \]

### Step 3: Use the relationship between A, G, and H

The relationship between the means is given by:
\[ G^2 = A \cdot H \]

Substituting \( A \) and \( G \) from the previous steps:
\[ (H + 1.6)^2 = (H + 3.6) \cdot H \]

### Step 4: Expand both sides

Expanding the left side:
\[ H^2 + 3.2H + 2.56 = H^2 + 3.6H \]

### Step 5: Simplify the equation

Now, we can simplify the equation by canceling \( H^2 \) from both sides:
\[ 3.2H + 2.56 = 3.6H \]

Rearranging gives:
\[ 2.56 = 3.6H - 3.2H \]
\[ 2.56 = 0.4H \]

### Step 6: Solve for H

Dividing both sides by 0.4:
\[ H = \frac{2.56}{0.4} = 6.4 \]

### Step 7: Find A and G

Now that we have \( H \), we can find \( A \) and \( G \):
1. For \( G \):
\[ G = H + 1.6 = 6.4 + 1.6 = 8 \]

2. For \( A \):
\[ A = H + 3.6 = 6.4 + 3.6 = 10 \]

### Step 8: Set up equations for a and b

We know:
1. \( A = \frac{a + b}{2} = 10 \) implies \( a + b = 20 \)
2. \( G = \sqrt{ab} = 8 \) implies \( ab = 8^2 = 64 \)

### Step 9: Solve the system of equations

Now we have a system of equations:
1. \( a + b = 20 \)
2. \( ab = 64 \)

Let’s solve for \( a \) and \( b \):
From \( a + b = 20 \), we can express \( b \) as:
\[ b = 20 - a \]

Substituting into the second equation:
\[ a(20 - a) = 64 \]
\[ 20a - a^2 = 64 \]
\[ a^2 - 20a + 64 = 0 \]

### Step 10: Use the quadratic formula

Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
\[ = \frac{20 \pm \sqrt{400 - 256}}{2} \]
\[ = \frac{20 \pm \sqrt{144}}{2} \]
\[ = \frac{20 \pm 12}{2} \]

### Step 11: Calculate the values of a and b

Calculating the two possible values:
1. \( a = \frac{32}{2} = 16 \)
2. \( b = \frac{8}{2} = 4 \)

Thus, the two numbers are \( a = 16 \) and \( b = 4 \).

### Final Answer:
The two numbers are 16 and 4.

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