Let p and q be roots of the equation x^ 2 - 2x + A = 0 and let r...

$x^{2} - 2x + A = 0$

$⇒ x = 1 \pm \sqrt{1 - A}$

$∴ p = 1 - \sqrt{1 - A}, q = 1 + \sqrt{1 - A}$

Similarly, $r = 9 - \sqrt{81 - B}, s = 9 + \sqrt{81 - B}$

$p, q, r, s$ are in $AP$.

$∵ q - p = s - r$

$⇒ 2\sqrt{1 - A} = 2\sqrt{81 - B}$

$⇒ B = 80 + A$

and $q - p = r - q$

$⇒ 3\sqrt{1 - A} = 8 - \sqrt{81 - B}$

$⇒ A = - 3, B = 77$