The first three terms of the sequence \( 6 ; x ; y ; 27 \) form an arithmetic progression and the last three terms form a geometric progression. Determine the values of \( x \) and \( y \). State the arithmetic and geometric sequences for each of your solutions.

To solve the problem, we begin by applying the definitions of arithmetic and geometric progressions. Since \( 6, x, y, 27 \) form an arithmetic progression, we can express the relationships between the terms using the property that the difference between consecutive terms is constant. Thus, we have: \[ x - 6 = y - x \quad \text{(1)} \] From this, we can rearrange to find: \[ 2x = 6 + y \quad \Rightarrow \quad y = 2x - 6 \quad \text{(2)} \] Next, since \( x, y, 27 \) form a geometric progression, we utilize the property that the ratio of consecutive terms is constant. Therefore: \[ \frac{y}{x} = \frac{27}{y} \quad \text{(3)} \] Cross multiplying gives: \[ y^2 = 27x \quad \Rightarrow \quad y = \sqrt{27x} \quad \text{(4)} \] Now we have two equations (2) and (4) that involve \( y \). By substituting equation (2) into equation (4): \[ 2x - 6 = \sqrt{27x} \] Next, we square both sides to eliminate the square root: \[ (2x - 6)^2 = 27x \] Expanding the left-hand side: \[ 4x^2 - 24x + 36 = 27x \] Rearranging this gives us: \[ 4x^2 - 51x + 36 = 0 \] Next, we use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{51 \pm \sqrt{(-51)^2 - 4 \cdot 4 \cdot 36}}{2 \cdot 4} \] \[ x = \frac{51 \pm \sqrt{2601 - 576}}{8} \] \[ x = \frac{51 \pm \sqrt{2025}}{8} \] \[ x = \frac{51 \pm 45}{8} \] Calculating the two possible values for \( x \): 1. \( x = \frac{96}{8} = 12 \) 2. \( x = \frac{6}{8} = \frac{3}{4} \) Now substituting back into equation (2) to find \( y \): For \( x = 12 \): \[ y = 2(12) - 6 = 24 - 6 = 18 \] For \( x = \frac{3}{4} \): \[ y = 2\left(\frac{3}{4}\right) - 6 = \frac{3}{2} - 6 = \frac{3}{2} - \frac{12}{2} = -\frac{9}{2} \] Now we check both solutions to see if they meet the conditions of the problem. 1. For \( (6, 12, 18, 27) \) as an arithmetic sequence: - Common difference: \( 12 - 6 = 6 \), \( 18 - 12 = 6 \), \( 27 - 18 = 9 \) - Common difference condition is met. As a geometric progression: - \( \frac{12}{6} = 2 \) - \( \frac{18}{12} = \frac{3}{2} \) - \( \frac{27}{18} = \frac{3}{2} \) - The ratio condition is consistent. Thus, \( x = 12 \) and \( y = 18 \) is valid. 2. For \( (6, \frac{3}{4}, -\frac{9}{2}, 27) \) as an arithmetic sequence: - Common difference calculations yield inconsistencies. Thus the only valid solution is: \[ \boxed{x = 12, y = 18} \] The arithmetic sequence is \( 6, 12, 18, 27 \) and the geometric sequence is \( 12, 18, 27 \).