Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}5G.\mathrm{M}.\mathrm{s}\mathrm{between}\frac{32}{9}\mathrm{and}\frac{81}{2}\mathrm{be}{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4\mathrm{and}{\mathrm{G}}_5. \\ \frac{32}{9},{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4,{\mathrm{G}}_5,\frac{81}{2} \\ \Rightarrow a=\frac{32}{9},n=7\mathrm{and}{a}_7=\frac{81}{2} \\ \because a_7=\frac{81}{2} \\ \Rightarrow a{r}^6=\frac{81}{2} \\ \Rightarrow r^6=\frac{81}{2}\times \frac{9}{32} \\ \Rightarrow r^6={\left(\frac{3}{2}\right)}^6 \\ \Rightarrow r=\frac{3}{2} \\ \therefore {\mathrm{G}}_1=a_2=ar=\frac{32}{9}\left(\frac{3}{2}\right)=\frac{16}{3} \\ {\mathrm{G}}_2=a_3=a{r}^2=\frac{32}{9}{\left(\frac{3}{2}\right)}^2=8 \\ {\mathrm{G}}_3=a_4=a{r}^3=\frac{32}{9}{\left(\frac{3}{2}\right)}^3=12 \\ {\mathrm{G}}_4=a_5=a{r}^4=\frac{32}{9}{\left(\frac{3}{2}\right)}^4=18 \\ {\mathrm{G}}_5=a_6=a{r}^5=\frac{32}{9}{\left(\frac{3}{2}\right)}^5=27 \end{gathered}$$