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Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

Since angle $A$ , $B$ and $C$ are angles of a triangle, we have $A+B+C = 180^\circ$ . And since they are in a arithmetic progression, we have $\color{#D61F06}{A+C=2B}$ $\implies A+B+C = 3B = 180^\circ$ $\implies B = 60^\circ$ . Therefore,

$\begin{aligned} Q & = \color{#3D99F6}{\frac ac} \sin 2C + \color{#3D99F6}{\frac ca} \sin 2A & \small \color{#3D99F6}{\text{By sine rule, }\frac ac = \frac{\sin A}{\sin C}} \\ & = \color{#3D99F6}{\frac {\sin A}{\sin C}}(2\sin C\cos C) + \color{#3D99F6}{\frac {\sin C}{\sin A}} (2\sin A\cos A) \\ & = 2(\sin A \cos C + \sin C \cos A) \\ & = 2 \sin (\color{#D61F06}{A+C}) \\ & = 2 \sin (\color{#D61F06}{2B}) \\ & = 2 \sin 120^\circ \\ & = \sqrt 3 \approx \boxed{1.732} \end{aligned}$

Take a 30-60-90 triangle and use definition of trigonometric functions.

As B=60 and A,B,C are in AP therefore we can assume that A=B=C=60(assuming common difference to be zero) therefore a=b=c(as all angels are 60 so equilatral triangle) then answer becomes Q=√3(putting A=C=60)=1.731