3. The first three terms of the sequence \( 6 ; x ; y ; 27 \) form an arithmetic progression and the last three terms form a geometric progression. 3.1 Determine the values of \( x \) and \( y \). 4.2 State the arithmetic and geometric sequences for each of your solutions.

To solve for \( x \) and \( y \) in the sequence \( 6; x; y; 27 \), we start with the conditions for arithmetic and geometric progressions. For the arithmetic progression \( 6, x, y, 27 \): The difference between consecutive terms is constant, so we have: \( x - 6 = y - x = 27 - y \). Let the common difference be \( d \). Then we can express \( x \) and \( y \) as: \( x = 6 + d \) and \( y = 6 + 2d \). Setting \( 27 - y = d \) gives: \( 27 - (6 + 2d) = d \) This leads to: \( 27 - 6 - 2d = d \) or \( 21 = 3d \), giving \( d = 7 \). Now substituting \( d = 7 \): \( x = 6 + 7 = 13 \) and \( y = 6 + 2(7) = 20 \). Next, for the geometric progression sequence \( x, y, 27 \): Here, the ratios between consecutive terms are equal: \(\frac{y}{x} = \frac{27}{y}\). Substituting \( x = 13 \) and \( y = 20 \) gives: \(\frac{20}{13} = \frac{27}{20}\) which is indeed true. Now, we state the sequences: Arithmetic: \( 6, 13, 20, 27 \) with a common difference of \( 7 \). Geometric: \( 13, 20, 27 \) with a common ratio \( \frac{20}{13} \) or approximately \( 1.538 \).