Question

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

Answer 1

The correct option is C

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

Explanation for the correct option

Step 1: Solve for common ratio

Given that $n$ geometric means be inserted between $a$ and $b$

Let the geometric means be $G_1,G_2,G_3,......,G_{n-1},G_n$

Then the series will be $a,G_1,G_2,G_3,......,G_{n-1},G_n,b$

$\therefore$Number of terms in the series is $N=n+2$

We know that the $n^{th}$ term of a G.P. is $t_N=a.r^{N-1}$ where $a$is the first term and $r$ is the common ratio.

For the given series,

$\begin{array}{crcl}& t_{n+2}& =& a{r}^{\left(n+2\right)-1}\\ \Rightarrow & b& =& a{r}^{\left(n+1\right)}\\ \Rightarrow & r^{\left(n+1\right)}& =& \frac{b}{a}\\ \Rightarrow & r& =& {\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}\end{array}$

Step 2: Solve for the required value

$\begin{array}{crcll}& t_{N-1}& =& a.r^{\left(N-1\right)-1}& \\ \Rightarrow & G_n& =& a.r^{\left(n+2-1\right)-1}& \left[\because N=n+2\right]\\ & & =& a.r^n& \\ \Rightarrow & G_n& =& a{\left(\frac{b}{a}\right)}^{\frac{n}{n+1}}& \left[\because r={\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}\right]\end{array}$

Hence the correct option is option(C) i.e. $a{\left(\frac{b}{a}\right)}^{\frac{n}{n+1}}$