4 The first three terms of the sequence \( 6 ; x ; y ; 16 \) form an arithmetic progression and the last three ferms form a geometric progression. 4.1 Determine \( x \) and \( y \). 4.2 State the arithmetic and geometric sequences for each of your solutions.

To find \( x \) and \( y \), we know that in an arithmetic progression, the difference between consecutive terms is consistent. Therefore, we can write the equation for \( x \) as: \[ x = 6 + d \] where \( d \) is the common difference. Then for \( y \): \[ y = x + d = (6 + d) + d = 6 + 2d \] Given that the last three terms \( x \), \( y \), and \( 16 \) form a geometric progression, the ratio between consecutive terms must be equal. Thus we have: \[ \frac{y}{x} = \frac{16}{y} \] Cross-multiplying gives: \[ y^2 = 16x \] Now substituting \( x \) and \( y \) in terms of \( d \): 1. From \( y = 6 + 2d \): \( (6 + 2d)^2 = 16(6 + d) \) Simplifying: \( 36 + 24d + 4d^2 = 96 + 16d \) Rearranging to form a quadratic equation: \( 4d^2 + 8d - 60 = 0 \) Dividing by 4: \( d^2 + 2d - 15 = 0 \) Factoring gives us: \( (d + 5)(d - 3) = 0 \) Thus, \( d = -5 \) or \( d = 3 \). 2. If \( d = 3 \): \( x = 6 + 3 = 9 \) \( y = 6 + 2(3) = 12 \) 3. If \( d = -5 \): \( x = 6 - 5 = 1 \) \( y = 6 + 2(-5) = -4 \) Thus, we have two sets of values for \( (x, y) \): \( (9, 12) \) and \( (1, -4) \). For the two cases: If \( (x, y) = (9, 12) \): - Arithmetic sequence: \( 6, 9, 12, 16 \) - Geometric sequence: \( 9, 12, 16 \) with common ratio \( \frac{12}{9} = \frac{16}{12} = \frac{4}{3} \). If \( (x, y) = (1, -4) \): - Arithmetic sequence: \( 6, 1, -4, 16 \) - Geometric sequence: \( 1, -4, 16 \) with common ratio \( \frac{-4}{1} = \frac{16}{-4} = -4 \).