$$\begin{gathered} If{x}_1,x_{3}aretherootsofa{x}^2-6x+1=0and{x}_2,x_{4}aretherootsof21x^2-bx+1=0 \\ and{x}_1,x_2,x_3,x_{4}are inH.P.then(a,b)= \end{gathered}$$

$\begin{array}{l}{x}_1,x_{3}arerootsofa{x}^2-6x+1=0then\\ x_1+x_3=\frac{6}{a},x_1{x}_3=\frac{1}{a}\\ x_2,x_{4}arerootsof21x^2-bx+1=0then\\ x_2+x_4=\frac{b}{21},x_2{x}_4=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$21$}\right.\\ 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(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}=\frac{1}{3}\\ and{x}_3=\frac{2x_2{x}_4}{x_2+x_4}=\frac{2\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$21$}\right.\right)}{\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$21$}\right.\right)}=\frac{2}{b}\\ \therefore x_2{x}_4=\frac{1}{21}\Rightarrow \frac{1}{3}\left(x_4\right)=\frac{1}{21}\Rightarrow x_4=\frac{1}{7}\\ \therefore x_2+x_4=\frac{b}{21}\\ \Rightarrow \frac{1}{3}+\frac{1}{7}=\frac{b}{21}\Rightarrow \frac{10}{21}=\frac{b}{21}\Rightarrow b=10\\ \therefore x_3=\frac{2}{10}=\frac{1}{5}\\ \therefore x_1{x}_3=\frac{1}{a}\\ \Rightarrow x_1=\frac{5}{a}\\ \Rightarrow x_1+x_3=\frac{6}{a}\Rightarrow \frac{5}{a}+\frac{1}{5}=\frac{6}{a}\Rightarrow \frac{6}{a}-\frac{5}{a}=\frac{1}{5}\\ \Rightarrow \frac{1}{a}=\frac{1}{5}\Rightarrow a=5\\ \therefore (a,b)=(5,10)\end{array}$