Official Solution:\(\{x, \ y, \ z\}\) is an increasing sequence of numbers such that the ratio between the consecutive terms is constant.
\(\{x + 2y, \ 2x + y + z, \ x + 3y + z\}\) is an increasing sequence of numbers such that the difference between the consecutive terms is constant.
What is the value of \(\frac{x}{z}\) ?A. \(\frac{1}{4}\)
B. \(\frac{1}{2}\)
C. \(1\)
D. \(2\)
E. \(4\)
The first statement implies that \(\frac{x}{y}=\frac{y}{z}\), which yields \(y^2 = xz\).
The second statement implies that \((2x + y + z) - (x + 2y) = (x + 3y + z) - (2x + y + z)\), which yields \(3y = z + 2x\).
Squaring \(3y = z + 2x\) yields \(9y^2= (z + 2x)^2\). Substituting \(y^2 = xz\) gives us \(9xz = (z + 2x)^2\).
\(9xz= z^2 + 4xz + 4x^2\)
\(4x^2 - 5xz + z^2 = 0\)
\(4x^2 - 4xz - xz + z^2 = 0\)
\(4x(x - z) - z(x - z) = 0\)
\((4x - z)(x - z)= 0\)
\(4x - z = 0\) or \(x - z = 0\), which is not possible since it would give \(x = z\), and this is not possible given that \(\{x, y, z\}\) is an
increasing sequence.
Therefore, \(4x = z\), and thus \(\frac{x}{z}=\frac{1}{4}\).
Answer: A