How do you find two geometric means between 5 and 135?

Question

21220 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-25T00:13:46

We are effectively looking for $a$ $a$ and $b$ $b$ such that $5$ $5$ , $a$ $a$ , $b$ $b$ , $135$ $135$ is a geometric sequence.

This sequence has common ratio $\sqrt[3]{\frac{135}{5}} = 3$ $root(3)(135/5) = 3$ , hence $a = 15$ $a = 15$ and $b = 45$ $b=45$

In a geometric sequence, each intermediate term is the geometric mean of the term before it and the term after it.

So we want to find $a$ $a$ and $b$ $b$ such that $5$ $5$ , $a$ $a$ , $b$ $b$ , $135$ $135$ is a geometric sequence.

If the common ratio is $r$ $r$ then:

$a = 5 r$ $a = 5r$

$b = a r = 5 r^{2}$ $b = ar = 5r^2$

$135 = b r = 5 r^{3}$ $135 = br = 5r^3$

Hence $r^{3} = \frac{135}{5} = 27$ $r^3 = 135/5 = 27$ , so $r = \sqrt[3]{27} = 3$ $r = root(3)(27) = 3$

Then $a = 5 r = 15$ $a = 5r = 15$ and $b = a r = 15 \cdot 3 = 45$ $b = ar = 15*3 =45$