Question

Insert 6 geometric means between 27 and $\frac{1}{81}$.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}6\mathrm{G}.\mathrm{M}.\mathrm{s}\mathrm{between}27\mathrm{and}\frac{1}{81}\mathrm{be}{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4,{\mathrm{G}}_5\mathrm{and}{\mathrm{G}}_6. \\ \mathrm{Thus},27,{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4,{\mathrm{G}}_5,{\mathrm{G}}_6\mathrm{and}\frac{1}{81}\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore a=27,n=8\mathrm{and}{a}_8=\frac{1}{81} \\ \because a_8=\frac{1}{81} \\ \Rightarrow {\mathrm{ar}}^7=\frac{1}{81} \\ \Rightarrow r^7=\frac{1}{81\times 27} \\ \Rightarrow r^7={\left(\frac{1}{3}\right)}^7 \\ \Rightarrow r=\frac{1}{3} \\ \therefore {\mathrm{G}}_1=a_2=ar=27\left(\frac{1}{3}\right)=9 \\ {\mathrm{G}}_2=a_3=a{r}^2=27{\left(\frac{1}{3}\right)}^2=3 \\ {\mathrm{G}}_3=a_4=a{r}^3=27{\left(\frac{1}{3}\right)}^3=1 \\ {\mathrm{G}}_4=a_5=a{r}^4=27{\left(\frac{1}{3}\right)}^4=\frac{1}{3} \\ {\mathrm{G}}_5=a_6=a{r}^5=27{\left(\frac{1}{3}\right)}^5=\frac{1}{9} \\ {\mathrm{G}}_6=a_7=a{r}^6=27{\left(\frac{1}{3}\right)}^6=\frac{1}{27} \end{gathered}$$