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I'm reading Sawyer's Prelude to Mathematics, from page 34-35:

[...]Prove that, if¹

$$\frac{ac-b^2}{a-2b+c}=\frac{bd-c^2}{b-2c+c^2}$$

Then the fractions just given are both equal to:

$$\frac{ad-bc}{a-b-c+d}$$

This question has a very definite form, and obviously to hammer it our by a lengthy and sharpless calculation, while veryfing the result, would bring one no nearer to the heart of the question. What interested me most was the question was the question, how did the examiner come to think of this question?

The pattern of the question includes the following aspects, $ac-b^2=0$ is the condition for the three quantities $a$, $b$, $c$ to be in geometrical progression.

1 - It is assumed that $b$ and $c$ are unequal. The text does not discuss this point, as it's not relevant to the main theme. What suggested the question to the examiner.

I've asked something similar before and someone said me it's the discriminant of the equation but here I'm in doubt on what he meant with conditon for $a$, $b$, $c$ to be in geometrical progression.

Later
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Red Banana
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1 Answers1

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We say that $a,b,c$ are in geometric progression if there is some $r$ such that $b=ra$ and $c=r^2a$, so the sequence reads $a,ra,r^2a$.

Note that unless $a=b=c=0$, $r$ must be $b/a$ so the condition for $a,b,c$ to be in geometric progression is $c=(b/a)^2a=b^2/a$, or equivalently that $b^2-ac=0$.

Alex Becker
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  • I can't understand where's the $ r$ coming from. – Red Banana Sep 18 '12 at 15:17
  • @GustavoBandeira I'm not sure what you mean. The existence of some such $r$ is the definition of being in geometric progression. – Alex Becker Sep 18 '12 at 19:58
  • I was thinking that I could switch any of the $a$, $b$, $c$ and it would always equals zero. Then I thought that this would be a little dumb, I guess it's impossible for it to hold true for all numbers. Your answer has some kind of heuristic that is still unknown to me. – Red Banana Sep 18 '12 at 20:14
  • @GustavoBandeira You mean if $b^2-ac=0$ then $a^2-bc=0$? Certainly not, consider $a=1,b=2,c=4$. – Alex Becker Sep 18 '12 at 20:16
  • I dunno how you proceed from the condition to the thing you made with the $r$. But it's starting to make sense. – Red Banana Sep 18 '12 at 20:18
  • Oh, you don't understand why $b^2-ac=0$ implies that there exists and $r$ such that $b=ra,c=r^2a$? – Alex Becker Sep 18 '12 at 20:21
  • I'm still stuck in the change the variables to numbers thing, I guess I need to process information only in an algebrical way, when I see an algebrical expression. – Red Banana Sep 18 '12 at 20:21
  • Yep. I understand it, but shouldn't the complete solution be: "the condition for $b^2-ac=0$ to be in geometrical progression is possible only if $b=ra$ and $c=r^2a$"? – Red Banana Sep 18 '12 at 20:24
  • That doesn't make sense. A sequence is in geometric progression, not an equation. So one would say "the condition for $a,b,c$ to be in geometric progression is that $b^2-ac=0$". The definition of $a,b,c$ being in geometric progression is that $b=ra$ and $c=r^2a$ for some $r$. – Alex Becker Sep 18 '12 at 20:31
  • At this question: Yes, I have no idea why it implies the existence of a $r$ such that $b=ra,c=r^2a$. – Red Banana Sep 27 '12 at 19:37
  • @GustavoBandeira "I have no idea why it" When you say "it", you mean "$b^2-ac=0$"? – Alex Becker Sep 27 '12 at 19:39
  • Yes. Just for curiosity, what else could it be? – Red Banana Sep 27 '12 at 19:47
  • @GustavoBandeira I don't know what else it could be. But if $b^2-ac=0$ then we can let $r=b/a$ and we get $b=(b/a)a=ra$ and $c=(c/a)a=(ac/a^2)a=(b^2/a^2)a=r^2a$. – Alex Becker Sep 27 '12 at 19:49