# Geometric progression related problem

1. Sep 4, 2013

### Saitama

1. The problem statement, all variables and given/known data
The sum of the squares of three distinct real numbers, which are in geometric progression is $S^2$. If their sum is $\alpha S$, show that $\alpha^2 \in (1/3,1) \cup (1,3)$.

2. Relevant equations

3. The attempt at a solution
Let the three numbers be $a , \, ar, \,ar^2$. According to the question
$$a^2+a^2r^2+a^2r^4=S^2 \, (*)$$
and
$$a+ar+ar^2=\alpha S \, (**)$$
Squaring (**), I get
$$a^2+a^2r^2+a^2r^4+2a^2r+2a^2r^3+2a^2r^2=\alpha^2S^2$$
From (*)
$$S^2+2a^2r(1+r+r^2)=\alpha^2S^2$$
The equation (**) can rewritten as $a(1+r+r^2)=\alpha S$. Using this
$$S^2+2ar(\alpha S)=\alpha^2S^2 \Rightarrow S+2a\alpha r=\alpha^2S$$
I am stuck here.

Any help is appreciated. Thanks!

2. Sep 4, 2013

### vela

Staff Emeritus
You have the two equations
\begin{align*}
\alpha S &= a(1+r+r^2) \\
S^2 &= a^2(1+r^2+r^4)
\end{align*} Can you see how you might solve for $\alpha^2$ while eliminating $S$ and $a$ at the same time?

3. Sep 5, 2013

### Tanya Sharma

After squaring eq (**) don't open the expression.Instead divide it by eq (*).This will eliminate 'S' and 'a' simultaneously.

Now,the tricky part .You can write
1+r2+r4 = (1+r+r2)(1-r+r2).

This will eliminate the expression (1+r+r2) from Nr and Dr.

You are now left with a quadratic equation in r2.Now in order to have a real value of 'r' the condition is ? I am sure you know this

Last edited: Sep 5, 2013
4. Sep 5, 2013

### Saitama

I get the following quadratic:
$$r^2(\alpha^2-1)-r(\alpha^2+1)+\alpha^2-1=0$$
I know I have to deal with discriminant here but do I have to use D>=0 or D>0 (D is discriminant)? I have tried using both of them and D>0 gives the correct answer but why not D>=0?

5. Sep 5, 2013

### vela

Staff Emeritus
You're supposed to be solving for $\alpha^2$, so you actually have a linear equation on your hands there. [strike]But how'd you get that? It's not correct.[/strike] Tanya basically told you how to do the whole problem.

Last edited: Sep 5, 2013
6. Sep 5, 2013

### Saitama

Yes I know I have to solve for $\alpha^2$ but Tanya also said that I would end up with quadratic in r. Is what I have written incorrect? From the quadratic, $\alpha^2 \neq 1$, the other values of $\alpha^2$ can be found out using the fact that D is greater than zero for the quadratic. My doubt is why not use D>=0? D>=0 gives 3 and 1/3 as possible solutions for $\alpha^2$ while the given answer does not include them. How am I supposed to know if I have to use D>=0 or D>0? I hope you understand my query.

7. Sep 5, 2013

### vela

Staff Emeritus
Sorry, your previous result was right. (I made the algebra mistake.)

You actually don't get a quadratic; you should get a rational expression. Try plotting your result for $\alpha^2$ as a function of r.

8. Sep 5, 2013

### Saitama

$\alpha^2=\frac{1+r+r^2}{1-r+r^2}$

I plotted this on Wolfram|Alpha: http://www.wolframalpha.com/input/?i=y=(1+x+x^2)/(1-x+x^2)
I have replaced $\alpha^2$ with y and r with x. It gives the result that range of y or $\alpha^2$ is [1/3,3] which is incorrect.

What's wrong with my method of evaluating the discriminant? I don't think the given problem requires me to do curve-sketching.

9. Sep 5, 2013

### vela

Staff Emeritus
You have to consider the allowed values of r. Not every value of r will satisfy the requirements stated in the first sentence of the problem. That will rule both of the D=0 roots.

From the graph, it's easy to see why two values of $\alpha^2$ should be excluded.

10. Sep 5, 2013

### Saitama

I think I understand it. r cannot be equal to 0,1 and -1 because for these values, the three terms won't be distinct. So 1,3 and 1/3 are excluded.

Thanks a lot vela and Tanya! :)

11. Sep 5, 2013

### vela

Staff Emeritus
You got it.

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