Question:

Step 1: Use the arithmetic progression condition to find angle B. Since angles $A$, $B$, and $C$ are in arithmetic progression, we have: $$2B = A + C$$ Combined with the angle sum property $A + B + C = 180°$, we can substitute: $$A + B + C = 180°$$ $$2B + B = 180°$$ $$3B = 180°$$ $$B = 60°$$ Step 2: Apply the Law of Cosines to relate the sides. Using the Law of Cosines: $$b^2 = a^2 + c^2 - 2ac\cos B$$ Substituting $B = 60°$ and $\cos 60° = \frac{1}{2}$: $$b^2 = a^2 + c^2 - 2ac \cdot \frac{1}{2}$$ $$b^2 = a^2 + c^2 - ac$$ Therefore: $$a^2 - ac + c^2 = b^2$$ Step 3: Simplify the given expression using the result from Step 2. The expression becomes: $$\frac{a+c}{\sqrt{a^2 - ac + c^2}} = \frac{a+c}{\sqrt{b^2}} = \frac{a+c}{b}$$ Step 4: Express $a + c$ using the Law of Sines. By the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$ where $R$ is the circumradius. Therefore: $$a + c = 2R(\sin A + \sin C)$$ Using the sum-to-product formula $\sin A + \sin C = 2\sin\frac{A+C}{2}\cos\frac{A-C}{2}$: $$a + c = 2R \cdot 2\sin\frac{A+C}{2}\cos\frac{A-C}{2}$$ Step 5: Express $b$ using the Law of Sines. $$b = 2R\sin B = 2R\sin 60° = 2R \cdot \frac{\sqrt{3}}{2} = R\sqrt{3}$$ Step 6: Simplify $\sin\frac{A+C}{2}$ using the angle sum property. Since $A + B + C = 180°$, we have $A + C = 180° - B = 180° - 60° = 120°$. Therefore: $$\sin\frac{A+C}{2} = \sin\frac{120°}{2} = \sin 60° = \frac{\sqrt{3}}{2}$$ Step 7: Calculate the final ratio. $$\frac{a+c}{b} = \frac{2R \cdot 2\sin\frac{A+C}{2}\cos\frac{A-C}{2}}{R\sqrt{3}}$$ $$= \frac{2R \cdot 2 \cdot \frac{\sqrt{3}}{2} \cdot \cos\frac{A-C}{2}}{R\sqrt{3}}$$ $$= \frac{2R\sqrt{3}\cos\frac{A-C}{2}}{R\sqrt{3}}$$ $$= 2\cos\frac{A-C}{2}$$ **Final Answer:** The value of $\dfrac{a+c}{\sqrt{a^2 - ac + c^2}}$ is $\boxed{2\cos\dfrac{A-C}{2}}$, which corresponds to **Option 4**.