The first three terms of the sequence \( 6 ; x ; y ; 16 \) form an arithmetic progression and the last three terms form a geometric progression. Determine \( x \) and \( y \). State the arithmetic and geometric sequences for each of your solutions.

To solve for \( x \) and \( y \), we start by using the properties of arithmetic and geometric progressions. For the arithmetic progression, the first three terms are \( 6, x, y \). We know that in an arithmetic progression, the difference between consecutive terms is constant. Therefore, we can express: \[ x - 6 = y - x \] This simplifies to: \[ 2x = 6 + y \quad (1) \] Rearranging gives us \( y = 2x - 6 \). Next, for the geometric progression formed by the last three terms \( x, y, 16 \), the ratio between consecutive terms must be constant. Thus, we have: \[ \frac{y}{x} = \frac{16}{y} \] Cross-multiplying leads to: \[ y^2 = 16x \quad (2) \] Now we can substitute equation (1) into equation (2). Replacing \( y \) in equation (2): \[ (2x - 6)^2 = 16x \] Expanding the left side gives: \[ 4x^2 - 24x + 36 = 16x \] Bringing everything to one side results in: \[ 4x^2 - 40x + 36 = 0 \] Dividing the entire equation by 4 simplifies to: \[ x^2 - 10x + 9 = 0 \] Now, we can factor or use the quadratic formula to find \( x \): \[ (x - 1)(x - 9) = 0 \] This gives us: \[ x = 1 \quad \text{or} \quad x = 9 \] **Case 1: \( x = 1 \)** Using \( x = 1 \) in (1): \[ y = 2(1) - 6 = 2 - 6 = -4 \] So one solution is \( (x, y) = (1, -4) \). **Case 2: \( x = 9 \)** Using \( x = 9 \) in (1): \[ y = 2(9) - 6 = 18 - 6 = 12 \] So another solution is \( (x, y) = (9, 12) \). Now, let's state the sequences: 1. For \( (x, y) = (1, -4) \): - Arithmetic sequence: \( 6, 1, -4 \) - Geometric sequence: \( 1, -4, 16 \) 2. For \( (x, y) = (9, 12) \): - Arithmetic sequence: \( 6, 9, 12 \) - Geometric sequence: \( 9, 12, 16 \) In conclusion, the values of \( (x, y) \) are \( (1, -4) \) and \( (9, 12) \), each forming their respective sequences as described above.