# If n geometric means between a and b be G_1,G_2,G_3,.......,G_n, and a geometric mean be G Then the value of G_1*G_2*G_3*........*G_n=(in terms of G)?

##### 1 Answer
Sep 22, 2017

Inserting $n$ geometric means between $a \mathmr{and} b$ we get the following GP series of total $\left(n + 2\right)$ terms

$a , {G}_{1} , {G}_{2} , {G}_{3} , \ldots \ldots \ldots . {G}_{n} , b$

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

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

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

Again G is the single GM between $a \mathmr{and} b$, So we have

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

$\implies a b = {G}^{2.} \ldots \ldots \left[2\right]$

And the product

${G}_{1} \times {G}_{2} \times {G}_{3} \times \ldots \ldots \ldots . \times {G}_{n}$

=prod_(i=1)^(i=n)G_i=prod_(i=1)^(i=n)ar^i=a^nr^(sum_(i=1)^(i=n)i=a^nr^((n(n+1))/2)

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

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

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

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

$= {\left({G}^{2}\right)}^{\frac{n}{2}}$ $\text{ "color(red)("Inserting } a b = {G}^{2}$

$= {G}^{n}$