# 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?

Nov 6, 2015

The ratio must be $r = \pm \sqrt{\frac{1 + \sqrt{5}}{2}}$

#### Explanation:

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

$a , a \cdot r , a \cdot {r}^{2}$.

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

${a}^{2} + {\left(a r\right)}^{2} = {\left(a {r}^{2}\right)}^{2}$

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

${x}_{1 , 2} = \frac{1 \setminus \pm \setminus \sqrt{5}}{2}$

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

Finally, we can solve for $r$:

${r}^{2} = x = \frac{1 + \sqrt{5}}{2} \setminus \implies r = \pm \sqrt{\frac{1 + \sqrt{5}}{2}}$