In a triangle, the lengths of the two larger sides are 10 and 9 respectively. If the angles are in A.P., the length of the third side can be:

$5-\sqrt{6}$ $3\sqrt{3}$ 5 $\sqrt{6}$

$5-\sqrt{6}$

2B = A + C ⇒ B = 60°. Also a = 10, b = 9   $∴ 2ac \cos B = (a^2 + c^2 - b^2)$ $⇒ 10c = (19 + c^2)$ or $c^2 - 10c +19 = 0 ⇒ c = 5 ± \sqrt{6}$ ∴ Since both the values of c are positive.