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08 Sep 2022, 04:55
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{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 x/z ?
A. 1/4
B. 1/2
C. 1
D. 2
E. 4
{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 x/z ?
A. 1/4
B. 1/2
C. 1
D. 2
E. 4
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30 Aug 2024, 09:50
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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
\(\{x, \ y, \ z\}\) 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 \(9xz = (z + 2x)^2\).
\(9xz= z^2 + 4xz + 4x^2\)
\(4x^2 - 5xz + z^2 = 0\)
\(4x^2 - 4xz - xz + z^2 = 0\)
\(4x(x - z) - z(x - z) = 0\)
\((4x - z)(x - z)= 0\)
Therefore, \(4x = z\), and thus \(\frac{x}{z}=\frac{1}{4}\).
Answer: A
General Discussion
08 Sep 2022, 06:20
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Bunuel
{x, y, z} is an increasing sequence of numbers such that the ratio between the consecutive terms is constant.
So, \(\frac{y}{x}=\frac{z}{y}.............xz=y^2...............\frac{xz}{z^2}=\frac{y^2}{z^2}............\frac{x}{z}=(\frac{y}{z})^2\)
{x + 2y, 2x + y + z, x + 3y + z} is an increasing sequence of numbers such that the difference between the consecutive terms is constant.
So, \(2x + y + z-(x + 2y)=x + 3y + z-(2x + y + z)..............x+z-y=2y-x............2x+z=3y\)
Divide by z
\(2(\frac{x}{z})+1=3(\frac{y}{z})\)
Substitute \(\frac{x}{z}=(\frac{y}{z})^2\)
\(2(\frac{y}{z})^2+1=3\frac{y}{z}\)
Let \(\frac{y}{z}=a\), then \(\frac{x}{z}=a^2\) and \(2a^2+1=3a.......2a^2-3a+1=0............(2a-1)(a-1)=0\)
So, \(a=\frac{1}{2}\) or a=1, but \(y\neq z\), so \(a^2=\frac{1}{4}\)
A
