Question

If x,y,z are the three geometric means between 6,54then z=

Answer 1

Geometric progression

Here, the first term is $a=6$ and the last term is $l=54$.

Number of terms when three geometric means are added between $a$ and $l$ is

$\quad\quad n=3+2=5$.

Here, $t_{5}=54$

$\Rightarrow ar^{4}=54$

$\Rightarrow 6.r^{4}=54$

$\Rightarrow r^{4}=9$

$\Rightarrow r=\pm\sqrt{3}$

Case 1. When $a=6,\:r=\sqrt{3}$

The three geometric means are

$\quad ar,\:ar^{2},\:ar^{3}$

$=6\sqrt{3},\:18,\:18\sqrt{3}$

So, $\color{blue}z=18\sqrt{3}$

Case 2. When $a=6,\:r=-\sqrt{3}$

The three geometric means are

$\quad ar,\:ar^{2},\:ar^{3}$

$=-6\sqrt{3},\:18,\:-18\sqrt{3}$

So, $\color{blue}z=-18\sqrt{3}$

AN EASIER WAY.

We have found: $a=6,\:r=\pm\sqrt{3}$

Here, $z = t_{4}$

$\Rightarrow z=a.r^{3}$

$\Rightarrow z=6.(\pm\sqrt{3})^{3}$

$\Rightarrow \color{blue}{z=\pm 18\sqrt{3}}$