 # Show that the product of n geometric means between a and b is equal

Question:

Show that the product of n geometric means between a and b is equal to the nth power of the single GM between a and b.

Solution:

To prove: Product of n geometric means between a and b is equal to the nth power of the single GM between a and b.

Formula used:(i) Geometric mean between $a$ and $b=\sqrt{a b}$

(ii) Sum of $n$ terms of A.P. $=\frac{(n)(n+1)}{2}$

Let the $\mathrm{n}$ geometric means between and $\mathrm{b}$ be $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \ldots \mathrm{G}_{n}$

Hence $a, G_{1}, G_{2}, G_{3}, \ldots . G_{n,} b$ are in GP

$\Rightarrow G_{1}=a r, G_{2}=a r^{2}$ and so on $\ldots$

Now, we have n+2 term

$\Rightarrow \mathrm{b}=\mathrm{ar}^{\mathrm{n}+2-1}$

$\Rightarrow \mathrm{b}=\mathrm{ar}^{\mathrm{n}+1}$

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

The product of $n$ geometric means is $G_{1} \times G_{2} \times G_{3} \times \ldots G_{n}$

$=a r \times a r^{2} \times a r^{3} \times \ldots a r^{n}$

$=a^{n} \times r^{(1+2+3 \ldots+n)}$

$=a^{n} \times r^{(n)\left(\frac{n+1}{2}\right)}\left[\right.$ Sum of $n$ terms of A.P. $\left.=\frac{(n)(n+1)}{2}\right]$

Substituting the value of r from eqn. (i)

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

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

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

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

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

$=(\sqrt{a b})^{n}$  … (ii)

Single geometric mean between a and b $=\sqrt{a b}$

$\mathrm{n}^{\mathrm{th}}$ power of single geometric mean between $\mathrm{a}$ and $\mathrm{b}=(\sqrt{\mathrm{ab}})^{\mathrm{n}}$

Hence Proved