If $x:y=2:1$ and $y:z=2:1$, then $x$,$y$,$z$ are continued proportional? and/or $z:x=1:4$? and/or $y^2+z x=4yz$?

Question

I will prove that I attempted this math by listing the things that I know.

1. $y=2z$
2. $x=2z^2$
3. $x=yz$

Also, did you notice something? If $x=8, y=4,z=2$ then everything works out.

I figured this while attempting the math.

But the problem is I don't know how to justify my hypothesis. That's why I'm asking the question.

If there's any problem in my question please inform me.

Answer 1

$x:y = 2:1$ so $x = 2y$.

$y:z = 2:1$ so $y = 2z$.

So $x = 2y = 2(2z) = 4z$ and so $x:z = 4: 1$ and $z:x = 1:4$.

So $ii$ is true.

Does $y^2 + xy = 4xy$? Well, $y^2 = (2z)^2 = 4z^2$ and $xy = (4z)(2z) = 8z^2$ so $y^2 + xy = 4z^2 + 8z^2 = 12z^2$. Meanwhile $4xy = 4(4z)(2z) = 32z^2$. Does $12z^2 = 32z^2$? Only if $z= 0$, which it can not as $y:z = 2:1$. So $iii$)

What about $i$? $x,y,z$ are quantuum porpoises[1]? I don't know what that means. But and $ii$ is true and $iii$ is false that means 1) is the only possible option.

1. $y = 2z$.

True.

2. $x = 2z^2$.

False.

3. $x = yz$

What the .....???????

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[1] Does "continued proportional" mean that $x:y = y:z$? Well, as $x:y = 2:1$ and $y:z = 2:1$ that is simply true. There's nothing to argue. It's just true.

Answer 2

From the condition we have that $x=2y; y=2z$. This yields that $x=4z$ and so ii) must be a true assertation. Now to check the validity of iii) just write everything in terms of $z$. You would get:

$$y^2 + zx = (2z)^2 + z\cdot 4z = 8z = 4 (2z)z = 4zy$$

Thus iii) is a true claim too. For the first claim continued proportional is a bit vague. If it means that the variables are ordered in increasing order then it's wrong. As you can take $x=-4, y=-2, z= -1$. Nevertheless we must have $x > y > z$ or $x < y < z$ depending on the sign of the variable.

Answer 3

\begin{align} x:y = 2:1, \ y:z = 2:1 &\implies \dfrac xy = 2, \ \dfrac yz = 2 \\ &\implies \dfrac xz = \dfrac xy \cdot \dfrac yz= 4\\ &\implies x:z=4:1 \\ \hline &\implies x:y = y:z \\ &\implies x:y :: y:z \\ \hline &\implies \dfrac xy + \dfrac yz = 4 \\ &\implies y^2 + zx = 4yz \end{align}

So all three are true.

Answer 4

Note that: $$x:y=2:1 \iff x:y=2:1=4:2.$$ It implies: $$x:y=4:2 \ \ \text{and} \ \ y:z=2:1 \Rightarrow x:y:z=4:2:1 \Rightarrow x=4n,y=2n,z=n.$$ Hence: $$\begin{align}i)& \ x:y:z=4:2:1 \ \ \text{(continued proportional)};\\ ii)& \ z:x=n:4n=1:4;\\ iii)& \ y^2+zx=4yz \iff (2n)^2+n(4n)=4(2n)n \iff 8n^2=8n^2. \end{align}$$ Hence, the correct answer is $4)$.