Q. If $n$ geometric means be inserted between $a$ and $b$, then the $n$th geometric mean will be

A) $a\left( \frac{b}{a} \right)^{\frac{n}{n - 1}}$
B) $a\left( \frac{b}{a} \right)^{\frac{n - 1}{n}}$
C) $a\left( \frac{b}{a} \right)^{\frac{n}{n + 1}}$
D) $a\left( \frac{b}{a} \right)^{\frac{1}{n}}$
Explanation

(c)
In $n$ geometric means $G_{1},\ G_{2},\ldots,G_{n}$ are to be inserted between two positive real numbers $a$ and $b$, then $a,\ G_{1},\ G_{2},\ldots,G_{n},$ $b$ are in GP, then
$G_{1} = ar,\ G_{2} = ar^{2},\ldots,\ G_{n} = ar^{n}$
So, $b = ar^{n + 1} \Rightarrow r = \left( \frac{b}{a} \right)^{\frac{1}{(n + 1)}}$
Now, $n$th geometric mean
$\left( G_{n} \right) = ar^{n} = a\left( \frac{a}{b} \right)^{\frac{n}{n + 1}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.