The lengths of the sides in a right triangle form 3 consecutive terms of a geometric sequence. How do you find the common ratio of the sequence?

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The lengths of the sides in a right triangle form 3 consecutive terms of a geometric sequence. How do you find the common ratio of the sequence?

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Answer 1 · 2015-11-06T10:45:31

The ratio must be $r=pmsqrt((1+sqrt(5))/2)$

Explanation:

Three numbers are in a geometric series if they're of the form

$a, a*r, a*r^2$ .

If they're sides of a right triangle, then we have

$a^2 + (ar)^2 = (ar^2)^2$

which is

$a^2 + a^2 r^2 = a^2 r^4$

Simplifying by $a^2$ :

$1+r^2=r^4$

To solve this equation (which of course you can rewrite as $r^4-r^2-1=0$ ) you can use a variable $x=r^2$ to turn it to a quadratic:

$x^2-x-1=0$

And this equation is quite famous, since it is the one whose solution involve the golden ratio, so

$x_{1,2} = (1\pm\sqrt(5))/2$

Now we have to remember that $x=r^2$ , and thus it must be positive. Since $(1+sqrt(5))/2$ is positive, and $(1-sqrt(5))/2$ is negative, we can only choose $x=(1+sqrt(5))/2$

Finally, we can solve for $r$ :

$r^2 = x = (1+sqrt(5))/2 \implies r=pmsqrt((1+sqrt(5))/2)$

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The lengths of the sides in a right triangle form 3 consecutive terms of a geometric sequence. How do you find the common ratio of the sequence?

1 Answer

Explanation:

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