Question

Prove that the product of $n$ geometric means between two positive numbers is the $n^{th}$ power of the single G.M. between them.

Answer 1

New common ratio

$r=\sqrt[n+1]{n+1\frac{b}{a}}$

G.M.s are $a{\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}$, $a{\left(\frac{b}{a}\right)}^{\frac{2}{n+1}}$, $a{\left(\frac{b}{a}\right)}^{\frac{3}{n+1}}$, .................................$a{\left(\frac{b}{a}\right)}^{\frac{n}{n+1}}$

Their product would be

$a{\left(\frac{b}{a}\right)}^{\frac{1+2+3+...........n}{n+1}}$

$=a{\left(\frac{b}{a}\right)}^{\frac{n(n+1)}{2(n+1)}}$

Single GM would be $a{\left(\frac{b}{a}\right)}^{\frac{1}{2}}$ whose nth power clearly equals our product.