The A.M. between two numbers is 20 and their G.M. is 16. find the numbers.

Answer

To solve the problem, we need to find two numbers \( a \) and \( b \) given that their Arithmetic Mean (A.M.) is 20 and their Geometric Mean (G.M.) is 16.

### Step-by-Step Solution:

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

2. **Setting Up the Equations**:
- From the problem, we know:
\[ \frac{a + b}{2} = 20 \]
Multiplying both sides by 2 gives:
\[ a + b = 40 \quad \text{(Equation 1)} \]
- For the G.M.:
\[ \sqrt{ab} = 16 \]
Squaring both sides gives:
\[ ab = 256 \quad \text{(Equation 2)} \]

3. **Expressing One Variable in Terms of the Other**:
- From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = \frac{256}{a} \]

4. **Substituting into the First Equation**:
- Substitute \( b \) from the above expression into Equation 1:
\[ a + \frac{256}{a} = 40 \]
- To eliminate the fraction, multiply through by \( a \):
\[ a^2 + 256 = 40a \]
- Rearranging gives:
\[ a^2 - 40a + 256 = 0 \quad \text{(Quadratic Equation)} \]

5. **Solving the Quadratic Equation**:
- We will use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( A = 1, B = -40, C = 256 \).
- Calculate the discriminant:
\[ D = (-40)^2 - 4 \cdot 1 \cdot 256 = 1600 - 1024 = 576 \]
- Now apply the quadratic formula:
\[ a = \frac{40 \pm \sqrt{576}}{2} \]
\[ \sqrt{576} = 24 \]
\[ a = \frac{40 \pm 24}{2} \]
- This gives two possible solutions:
1. \( a = \frac{64}{2} = 32 \)
2. \( a = \frac{16}{2} = 8 \)

6. **Finding the Corresponding Values of \( b \)**:
- If \( a = 32 \):
\[ b = \frac{256}{32} = 8 \]
- If \( a = 8 \):
\[ b = \frac{256}{8} = 32 \]

7. **Final Answer**:
- The two numbers are \( 32 \) and \( 8 \).