AP and their ratios

$\dfrac {A_1}{B_1} = \dfrac {a_1}{b_1} = \dfrac {1-1}2 = 0 \implies a_1 = 0$ . Let the common difference of $\{A_n\}$ and $\{B_n \}$ be $d_a$ and $d_b$ respectively. Then

$\begin{cases} \dfrac {A_2}{B_2} = \dfrac {d_a}{2b_1+d_b} = \dfrac 14 & \implies 4 d_a = 2b_1 + d_b & ...(1) \\ \dfrac {A_3}{B_3} = \dfrac {2d_a}{2b_1+2d_b} = \dfrac 13 & \implies 3 d_a = b_1 + d_b & ...(2) \end{cases}$

From $(1) - (2)$ : $d_a = b_1$ and from $(1)$ : $d_b = 2b_1$ . Therefore $a_n = (n-1) b_1$ and $b_n = b_1 + 2b_1(n-1) = (2n-1)b_1$ and

$\begin{aligned} \frac {a_3+a_5+a_7}{3(b_3+b_9)} + \frac {a_4+a_{10})}{2(b_2+b_{10}} & = \frac {2b_1+4b_1+6b_1}{3(5b_1+17 b_1)} + \frac {3b_1+9b_1}{2(3b_1+19b_1)} = \frac {12}{3(22)} + \frac {12}{2(22)} = \boxed {\frac 5{11}}\end{aligned}$

Use the fact that $A_n,B_n$ are both (as sums of arithmetic progressions) quadratic functions of $n$ , and that $A_0=B_0=0$ to find that $A_n=k \left( n^2-n \right)$ and $B_n=2kn^2$ , where $k$ is an arbitrary non-zero constant.

Now use the fact that $a_n=A_n-A_{n-1}$ to find $a_n=2k(n-1)$ and similarly $b_n=2k(2n-1)$ .

It's easy now to substitute in and find the required fraction $\boxed{\frac{5}{11}}$ .