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

The angles of the triangle are in A.P.
Now one angle is greater that 60° and other is less than 60°. Let A>60°.
Thus, larger angles are A and B
$\therefore \mathrm{a}=10\text{ and }\mathrm{b}=9;\mathrm{\angle }\mathrm{B}={60}^{\circ }$
Using the cosine formula $\cos {60}^{\circ }=\frac{100+{\mathrm{c}}^2-81}{2\times 10\times \mathrm{c}}$
$\Rightarrow {\mathrm{c}}^2-10\mathrm{c}+19=0\Rightarrow \mathrm{c}=5\pm \sqrt{6}$
Given that a = 10, b = 9 are the longer sides.
$\therefore \mathrm{c}=5\pm \sqrt{6}$