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
KillerBunny
Nov 6, 2015

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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