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. $a{\left(\frac{b}{a}\right)}^{\frac{n-1}{\left(n\right)}}$
• C. $a{\left(\frac{b}{a}\right)}^{\frac{n}{(n+1)}}$
• D. $a{\left(\frac{b}{a}\right)}^{\frac{1}{\left(n\right)}}$

Answer 1

The correct option is C $a{\left(\frac{b}{a}\right)}^{\frac{n}{(n+1)}}$

If n geometric means $g_1,g_2.....g_n$ are to be inserted between two positive real numbers a and b, then a, $g_1,g_2......g_n,$ b are in G.P. Then
$g_1=ar,g_2=a{r}^2......g_n=a{r}^n$
So $b=a{r}^{n+1}\Rightarrow r={\left(\frac{b}{a}\right)}^{1/(n+1)}$
Now $n^{th}$ geometric mean $\left(g_n\right)=a{r}^n=a{\left(\frac{b}{a}\right)}^{n/(n+1)}.$
Aliter : As we have the $m^{th}$ G.M. is given by $\left(G_m\right)=a{\left(\frac{b}{a}\right)}^{\frac{n}{(n+1)}}$
Now replace m by n we get the required result.