Given that $\alpha ,\gamma$ be the roots of $A{x}^2-4x+1=0$ and $\beta ,\delta$ the roots of $B{x}^2-6x+1=0$, then the values of A and B such that $\alpha ,\beta ,\gamma ,\delta$ are in HP.

Given that $\alpha ,\gamma$ are the roots of $A{x}^2-4x+1=0$

$\Rightarrow \alpha +\gamma =\frac{4}{A}$ and $\alpha \gamma =\frac{1}{A}$  ... (i)

$\frac{1}{\alpha }+\frac{1}{\gamma }=4$ Also, $\beta$ and $\delta$ are the roots of $B{x}^2-6x+1=0$.

$\Rightarrow \beta +\delta =\frac{6}{B}$ and $\beta \delta =\frac{1}{B}$  ... (ii)

$\frac{1}{\beta }+\frac{1}{\delta }=6$

Also, it is given that $\alpha ,\beta ,\gamma ,\delta \in$HP

$\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }\in \mathrm{AP}$

Now, from Eq. (i), we get

$\begin{array}{l}\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\gamma }}=4\\ \Rightarrow \frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\alpha }}+2\mathrm{d}=4\\ \Rightarrow \frac{2}{\mathrm{\alpha }}+2\mathrm{d}=4\\ \Rightarrow \frac{1}{\mathrm{\alpha }}+\mathrm{d}=2 ... \left(\mathrm{iii}\right)\end{array}$

Again, $\frac{1}{\beta }+\frac{1}{\delta }=6$

$\begin{array}{l}\Rightarrow \frac{1}{\alpha }+d+\frac{1}{\alpha }+3d=6\\ \Rightarrow \frac{2}{\alpha }+4d=6\\ \Rightarrow \frac{1}{\alpha }+2d=3 ... \left(\mathrm{iv}\right)\end{array}$

Solving Eqs (iii) and (iv), we get

$\alpha =1,\delta =1$

Now $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }=\frac{1}{\alpha },\frac{1}{\alpha }+d,\frac{1}{\alpha }+2d,\frac{1}{\alpha }+3d=\frac{1}{1},\frac{1}{1}+1,\frac{1}{1}+2.1,\frac{1}{1}+3.1$

$=1,2,3,4$

Thus $\alpha =1,\beta =\frac{1}{2},\gamma =\frac{1}{3},\delta =\frac{1}{4}$

Therefore, $A=\frac{1}{\alpha \gamma }=3,B=\frac{1}{\beta \delta }=8$