Let the four numbers in $A.P.$ be

$p=a-3d, q=a-d, r=a+d, s=a+3d$.

Now, $p$ and $q$ are the roots of the quadratic equation $x^2-2x+A=0$.

Therefore, using the sum and product of roots of a quadratic equation, we get

$p+q=2 \dots \left(i\right)$

$pq=A \dots \left(ii\right)$

And, $r,s$ are the roots of the quadratic equation $x^2-18x+B=0$.

Therefore,

$r+s=18 \dots \left(iii\right)$

$rs=B\mathit{ }\mathit{ }\mathit{ }\mathit{\dots }\left(iv\right)$

By $\left(i\right)$ and $\left(iii\right)$, we have

$p+q+r+s=4a=20$

$\Rightarrow a=5$

Now,

$p+q=2$

$\Rightarrow 10-4d=2$

$\Rightarrow d=2.$

Also, 

$r+s=18\begin{array}{}\end{array}$

$\Rightarrow 10+4d=18$

$\Rightarrow d=2\begin{array}{}\end{array}.$

Hence, the numbers are $-1, 3, 7, 11.$

Therefore, $pq=A=-3, rs=B=77.$