Question

If $n$ geometric means be inserted between $a$ and $b$, then prove that their products is ${\left(ab\right)}^{n/2}$.

Answer 1

Inserting $n$ geometric means between $a$ and $b$ we get the following GP series of total $(n+2)$ terms

$a,G_1,G_2,G_3,..........G_n,b$

Let $r$ be the common ratio of this GP then $b$ becomes the $n+2$ th term of the series. So we have,

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

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

Again $G$ is the single GM between $a$ and $b$, So we have

$G=(ab{)}^{\left(\frac{1}{2}\right)}$

$⟹ab=G^2$.......(2)

And the product

$G_1\times G_2\times G_3\times ..........\times G_n$

$={\prod }_{i=1}^{i=n}{G}_i={\prod }_{i=1}^{i=n}a{r}^i=a^n{r}^{{\sum }_{i=1}^{i=n}i}=a^n{r}^{\frac{n(n+1)}{2}}$

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

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

$=(a{)}^{\frac{n}{2}}\times (b{)}^{\frac{n}{2}}$

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