Since, A, B, C are in AP
$$\Rightarrow$$ 2B = A + C i.e., $$\angle$$B = 60$$^\circ$$
$$\therefore$$ $${a \over c}$$(2 sin C cos C) + $${c \over a}$$ (2 sin A cos A)
= 2k (a cos C + c cos A)
[using, $${a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = {1 \over k}$$]
= 2k (b)
= 2 sin B
[using, b = a cos C + c cos A]
= $$\sqrt3$$
