If H.M. of two number is 4 , then A.M. 'A' and G.M. 'G' satisfy the relation ` 2A + G^(2) = 27` , then modulus of
difference of these two numbers is

Answer

To solve the problem, we need to find the modulus of the difference between two numbers \( x \) and \( y \) given that their Harmonic Mean (H.M.) is 4 and they satisfy the relation \( 2A + G^2 = 27 \), where \( A \) is the Arithmetic Mean (A.M.) and \( G \) is the Geometric Mean (G.M.).

### Step-by-Step Solution:

1. **Understanding H.M.**:
The formula for the Harmonic Mean of two numbers \( x \) and \( y \) is given by:
\[ H.M. = \frac{2xy}{x+y} \]
Given that \( H.M. = 4 \), we can set up the equation:
\[ \frac{2xy}{x+y} = 4 \]

2. **Cross-multiplying**:
Cross-multiplying gives us:
\[ 2xy = 4(x+y) \]
Simplifying this, we have:
\[ 2xy = 4x + 4y \]
Dividing through by 2:
\[ xy = 2x + 2y \quad \text{(1)} \]

3. **Understanding A.M. and G.M.**:
The Arithmetic Mean \( A \) and Geometric Mean \( G \) are given by:
\[ A = \frac{x+y}{2} \quad \text{and} \quad G = \sqrt{xy} \]
Therefore, we can express \( A \) and \( G^2 \):
\[ G^2 = xy \]

4. **Substituting into the given relation**:
We know from the problem that:
\[ 2A + G^2 = 27 \]
Substituting \( A \) and \( G^2 \):
\[ 2\left(\frac{x+y}{2}\right) + xy = 27 \]
This simplifies to:
\[ x + y + xy = 27 \quad \text{(2)} \]

5. **Using equations (1) and (2)**:
From equation (1), we have \( xy = 2x + 2y \). We can substitute this into equation (2):
\[ x + y + (2x + 2y) = 27 \]
Simplifying gives:
\[ 3x + 3y = 27 \]
Dividing through by 3:
\[ x + y = 9 \quad \text{(3)} \]

6. **Finding \( xy \)**:
Now substituting \( x + y = 9 \) back into equation (1):
\[ xy = 2(x + y) = 2 \times 9 = 18 \quad \text{(4)} \]

7. **Forming a quadratic equation**:
We now have:
\[ x + y = 9 \quad \text{and} \quad xy = 18 \]
The numbers \( x \) and \( y \) are the roots of the quadratic equation:
\[ t^2 - (x+y)t + xy = 0 \]
Substituting the values:
\[ t^2 - 9t + 18 = 0 \]

8. **Solving the quadratic equation**:
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ t = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2} \]
This gives us:
\[ t = \frac{12}{2} = 6 \quad \text{or} \quad t = \frac{6}{2} = 3 \]
Thus, \( x = 6 \) and \( y = 3 \) (or vice versa).

9. **Finding the modulus of the difference**:
The modulus of the difference is:
\[ |x - y| = |6 - 3| = 3 \]

### Final Answer:
The modulus of the difference of the two numbers is \( \boxed{3} \).