Question

Prove that the product of $n$ geometric mean between any two numbers is $nth$ power of their $G.M$.

Answer 1

Let a and b be the 2 numbers and G be the geometric mean between them.

Therefore a,b,G must be in G.P.

Common ratio $=\frac{G}{a}=\frac{b}{G}$

$\Rightarrow G^2=ab\to G=\sqrt{ab}$...(1)

Therefore,

$a,G_1,G_2,......,G_n,b$ form G.P.

This series has a as its first term and b as last and (n+2)th term

$\therefore b=a{r}^{n+2-1}$

$\Rightarrow b=a{r}^{n+1}$

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

Product $=G_1{G}_2...G_n$

$=\left(ar\right)\left(a{r}^2\right).....\left(a{r}^n\right)$

$=a^n{r}^{1+2+..+n}$

$=a^n{r}^{\frac{n(n+1)}{2}}$

From (2) we get

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

$=(ab{)}^{\frac{n}{2}}$

$=(\sqrt{ab}{)}^n$

From (1) we get

Product $=G^n$

$G_1{G}_2....G_n=G^n$