Question

Insert 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}$.

Answer 1

Let $G_1,G_2,G_3,G_4,G_5$ be 5 geometric means between $\frac{32}{9}$ and $\frac{81}{2}$

Then, $\frac{32}{9},G_1,G_2,G_3,G_4,G_5,\frac{81}{2}$ is a G.P.

with $a=\frac{32}{9},b=\frac{81}{2}$

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

$={\left(\frac{\frac{81}{2}}{\frac{32}{9}}\right)}^{\frac{1}{5+1}}={(\frac{81}{2}\times \frac{9}{32})}^{\frac{1}{6}}=\left(\frac{3^6}{2^6}\right)=\frac{3}{2}$

Thus, $G_1=ar=\frac{32}{9}\times \frac{3}{2}=\frac{16}{3}$

$G_2=a{r}^2=\frac{32}{9}\times \frac{9}{4}=8$

$G_3=a{r}^3=\frac{32}{9}\times \frac{27}{8}=12$

$G_4=a{r}^4=\frac{32}{9}\times \frac{3^4}{2^4}=2times9=18$

$G_5=a{r}^5=\frac{32}{9}\times \frac{3^5}{2^5}=27$

Hence, $\frac{16}{3},8,12,18,27$ are 5 geometric means

between $\frac{32}{9}$ and $\frac{81}{2}$.