Question

Insert 5 geometric means between 16 and $\frac{1}{4}$.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}5\mathrm{G}.\mathrm{M}.\mathrm{s}\mathrm{betweem}16\mathrm{and}\frac{1}{4}\mathrm{be}{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4\mathrm{and}{\mathrm{G}}_5. \\ 16,{\mathrm{G}}_1,{\mathrm{G}}_2,{\mathrm{G}}_3,{\mathrm{G}}_4,{\mathrm{G}}_5,\frac{1}{4} \\ \Rightarrow a=16,n=7\mathrm{and}{a}_7=\frac{1}{4} \\ \because a_7=\frac{1}{4} \\ \Rightarrow a{r}^6=\frac{1}{4} \\ \Rightarrow r^6=\frac{1}{4\times 16} \\ \Rightarrow r^6={\left(\frac{1}{2}\right)}^6 \\ \Rightarrow r=\frac{1}{2} \\ \therefore {\mathrm{G}}_1=a_2=ar=16\left(\frac{1}{2}\right)=8 \\ {\mathrm{G}}_2=a_3=a{r}^2=16{\left(\frac{1}{2}\right)}^2=4 \\ {\mathrm{G}}_3=a_4=a{r}^3=16{\left(\frac{1}{2}\right)}^3=2 \\ {\mathrm{G}}_4=a_5=a{r}^4=16{\left(\frac{1}{2}\right)}^4=1 \\ {\mathrm{G}}_5=a_6=a{r}^5=16{\left(\frac{1}{2}\right)}^5=\frac{1}{2} \end{gathered}$$