${\mathrm{x}}_1$ and ${\mathrm{x}}_3$ are the roots of the equation ${\mathrm{Ax}}^2-4\mathrm{x}+1=0$ and ${\mathrm{x}}_2$ and ${\mathrm{x}}_4$ are the roots of the equation ${\mathrm{Bx}}^2-6\mathrm{x}+1=0.$ If ${\mathrm{x}}_1, {\mathrm{x}}_2, {\mathrm{x}}_3, {\mathrm{x}}_4$ form a HP, then 

$\begin{array}{l}{x}_1,x_{3}arerootsofA{x}^2-4x+1=0then\\ x_1+x_3=\frac{4}{A},x_1{x}_2=\frac{1}{A}\\ and{x}_2,x_{4}arerootsofB{x}^2-6x+1then\\ x_2+x_4=\frac{6}{B},x_2{x}_4=\frac{1}{B}\\ If{x}_1,x_2,x_3,x_{4}areinH.P.then\\ x_2=\frac{2x_1{x}_3}{x_1+x_3}=\frac{2\left(\frac{1}{A}\right)}{\left(\frac{4}{A}\right)}=\frac{1}{2}\\ and{x}_3=\frac{2x_2{x}_4}{x_2+x_4}=\frac{2\left(\frac{1}{B}\right)}{\frac{6}{B}}=\frac{1}{3}\\ \therefore x_1{x}_3=\frac{1}{A}\Rightarrow x_1\left(\frac{1}{3}\right)=\frac{1}{A}\Rightarrow x_1=\frac{3}{A}\\ \therefore x_1+x_3=\frac{4}{A}\Rightarrow \frac{3}{A}+\frac{1}{3}=\frac{4}{A}\\ \Rightarrow \frac{1}{3}=\frac{1}{A}\Rightarrow A=3\\ x_2{x}_4=\frac{1}{B}\Rightarrow \frac{1}{2}\left(x_4\right)=\frac{1}{B}\\ \Rightarrow x_4=\frac{2}{B}\\ \therefore x_2+x_4=\frac{6}{B}\Rightarrow \frac{1}{2}+\frac{2}{B}=\frac{6}{B}\\ \Rightarrow \frac{1}{2}=\frac{4}{B}\Rightarrow B=8\\ \therefore (A,B)=(3,8)\end{array}$