Question:

Step 1: Establish the relationship between individual terms and the sum ratio. For arithmetic progressions, we can use the property that relates individual terms to the sum of first $n$ terms. The middle term of the first $2n-1$ terms equals the average, so: $$\frac{a_n}{b_n} = \frac{A_{2n-1}}{B_{2n-1}}$$ Substituting the given ratio $\dfrac{A_n}{B_n} = \dfrac{n-1}{2n}$: $$\frac{a_n}{b_n} = \frac{A_{2n-1}}{B_{2n-1}} = \frac{(2n-1)-1}{2(2n-1)} = \frac{2n-2}{4n-2} = \frac{n-1}{2n-1}$$ Step 2: Simplify the first fraction using arithmetic progression properties. For the first fraction $\dfrac{a_3+a_5+a_7}{3(b_3+b_9)}$: Using the AP property that the sum of terms equidistant from the middle equals twice the middle term: - $a_3 + a_5 + a_7 = 3a_5$ (since $a_5$ is the middle term) - $b_3 + b_9 = 2b_6$ (since $b_6$ is the average of $b_3$ and $b_9$) Therefore: $$\frac{a_3+a_5+a_7}{3(b_3+b_9)} = \frac{3a_5}{3 \cdot 2b_6} = \frac{a_5}{2b_6}$$ Step 3: Simplify the second fraction using arithmetic progression properties. For the second fraction $\dfrac{a_4+a_{10}}{2(b_2+b_{10})}$: Using the AP property: - $a_4 + a_{10} = 2a_7$ (since $a_7$ is the average of $a_4$ and $a_{10}$) - $b_2 + b_{10} = 2b_6$ (since $b_6$ is the average of $b_2$ and $b_{10}$) Therefore: $$\frac{a_4+a_{10}}{2(b_2+b_{10})} = \frac{2a_7}{2 \cdot 2b_6} = \frac{a_7}{2b_6}$$ Step 4: Combine both fractions and simplify. Adding the two fractions: $$\frac{a_5}{2b_6} + \frac{a_7}{2b_6} = \frac{a_5+a_7}{2b_6}$$ Using the AP property again, $a_5 + a_7 = 2a_6$: $$\frac{a_5+a_7}{2b_6} = \frac{2a_6}{2b_6} = \frac{a_6}{b_6}$$ Step 5: Calculate the final answer using the formula from Step 1. Using the relationship $\dfrac{a_n}{b_n} = \dfrac{n-1}{2n-1}$ with $n=6$: $$\frac{a_6}{b_6} = \frac{6-1}{2(6)-1} = \frac{5}{11}$$ **Final Answer:** $\boxed{\dfrac{5}{11}}$ This corresponds to **Option 4**.