Step 1: Define Given Relations We are given: 1. Arithmetic Mean (A): \[ A = \frac{1}{2} (x + y) \] 2. Geometric Mean (G): \[ G = \sqrt{xy} \quad \Rightarrow \quad G^2 = xy \] 3. Harmonic Mean (H) condition: \[ \frac{2xy}{x+y} = 4 \] 4. Sum condition: \[ 2A + G^2 = 27 \] Step 2: Solve for \( A \) and \( xy \) From the harmonic mean condition: \[ \frac{2xy}{x+y} = 4 \] Multiplying both sides by \( (x + y) \): \[ 2xy = 4(x + y) \] \[ xy = 2(x + y) \] Using the sum condition: \[ 2A + G^2 = 27 \] Substituting \( A = \frac{x+y}{2} \) and \( G^2 = xy \): \[ 2 \times \frac{x+y}{2} + xy = 27 \] \[ (x+y) + xy = 27 \] Substituting \( xy = 2(x+y) \): \[ (x+y) + 2(x+y) = 27 \] \[ 3(x+y) = 27 \] \[ x+y = 9 \] Substituting into \( xy = 2(x+y) \): \[ xy = 2(9) = 18 \] --- Step 3: Solve for \( x \) and \( y \) We now have: \[ x + y = 9, \quad xy = 18 \] These satisfy the quadratic equation: \[ t^2 - (x+y)t + xy = 0 \] \[ t^2 - 9t + 18 = 0 \] Factoring: \[ (t - 6)(t - 3) = 0 \] \[ t = 6 \quad \text{or} \quad t = 3 \] Thus, the numbers are 6 and 3. \[ \boxed{6, 3} \]