Question

Show that the geometric mean between x and y is ±√xy and the common ratio r is r=^n+1√y/a

Answer 1

If three quantities are in Geometric Progression then the middle one is called the geometric mean of the other two.

Let, three numbers $x, G$ and $y$ are in Geometric Progression then, the middle number $G$ is called the geometric mean between two numbers $x$ and $y.$

$x, G$ and $y$ are in Geometric Progression $G\not=0, xy>0$

$<=>\dfrac{G}{x}=\dfrac{y}{G}$

$<=>xy=G^2$

$G=\pm\sqrt{xy}$

Insert $n$ geometric means between $a$ and $y.$

Let $a_1, a_2, ..., a_n$ be $n$ geometric means between $a$ and $y.$

The numbers $a,a_1, a_2, ..., a_n, y$ are in Geometric Progression.

Common ratio

$$r={a_1\over a}$$

Then

$$y=ar^{n+1}$$

Hence

$$r=\sqrt[n+1]{{y\over a}}$$