Prove that the product of n geometric means between two quantities
Question:

Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.

Solution:

Let $G_{1}, G_{2}, G_{3}, G_{4}, \ldots, G_{n}$ be $n$ G. M.s between $a$ and $b$.

Then, $a, G_{1}, G_{2}, G_{3}, G_{4}, \ldots, G_{n}, b$ is a G.P.

Let $r$ be the common ratio.

$\because b=a_{n+2}=a r^{(n+1)}$

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

$\therefore G_{1}=a_{2}=a r$

$G_{2}=a_{3}=a r^{2}$

$G_{3}=a_{4}=a r^{3}$

$G_{n}=a_{(n+1)}=a r^{n}$

Also, let $G$ be the G.M. between $a$ and $b$.

$\therefore G^{2}=a b$

Now, $G_{1} \times G_{2} \times G_{3} \times G_{4} \times \ldots \times G_{n}=a r \times a r^{2} \times a r^{3} \times a r^{4} \times \ldots \times a r^{n}$

$=a^{n} \times r^{(1+2+3+4+\ldots \ldots+n)}$

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

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

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

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

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

$=(\sqrt{a b})^{n}$

$=G^{n}$

$\therefore G_{1} \times G_{2} \times G_{3} \times G_{4} \times \ldots \times G_{n}=G^{n}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.