Question

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?

Answer 1

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)`