$$\begin{gathered} \text{ Let }{\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3,\dots {\mathrm{A}}_{\mathrm{n}}\text{ be }\mathrm{n}\text{ arithmetic means between }20\text{ and }80 \\ \text{ And let }\mathrm{d}\text{ be the common difference between the terms of the A.P. } \\ \text{ So, }d=\frac{b-a}{n+1} \\ =\frac{80-20}{\mathrm{n}+1}=\frac{60}{\mathrm{n}+1} \\ \text{ Now, }{\mathrm{A}}_1=20+\mathrm{d} \\ =20+\frac{60}{\mathrm{n}+1}=20\left(\frac{\mathrm{n}+4}{\mathrm{n}+1}\right) \\ \text{ And }{\mathrm{A}}_{\mathrm{n}}=20+\mathrm{nd}=20+\frac{60\mathrm{n}}{\mathrm{n}+1}=20\left(\frac{4\mathrm{n}+1}{\mathrm{n}+1}\right) \\ \text{ Thus, }\frac{{\mathrm{A}}_1}{{\mathrm{A}}_{\mathrm{n}}}=\frac{1}{3}\text{ (given) } \\ \frac{20\left(\frac{\mathrm{n}+4}{\mathrm{n}+1}\right)}{20\left(\frac{4\mathrm{n}+1}{\mathrm{n}+1}\right)}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{n}+4}{4\mathrm{n}+1}=\frac{1}{3} \\ \therefore \text{ The value of }\mathrm{n}\text{ is }11 \end{gathered}$$