$(a)\;\large\frac{1}{2}$$\qquad(b)\;\large\frac{\sqrt 3}{2}$$\qquad(c)\;1\qquad(d)\;\sqrt 3$

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Since $A,B,C$ are in AP

$\Rightarrow 2B=A+C$

(i.e) $\sqrt B=60^{\large\circ}$

$\therefore \large\frac{a}{c}$$(2\sin C\cos C)+\large\frac{C}{a}$$(2\sin A\cos A)=2k(a\cos C+C\cos A)$

$\big[Using\;\large\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}=\frac{1}{k}\big]$

$\Rightarrow 2k(b)$

[Using $b=a\cos C+c\cos A$]

$\Rightarrow 2\sin B$

$\Rightarrow \sqrt 3$

Hence (d) is the correct answer.

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