Inspired by Ayush Rai

Let the first terms and common differences be $a_1$ and $a_2$ , and $d_1$ and $d_2$ respectively.

From $\dfrac {7n+1}{4n+27}$ for $n=1$ , we have $\dfrac {a_1}{a_2} = \dfrac 8{31}$ $\implies a_1 = \dfrac 8{31}a_2$ .

We note that the nominator $7n+1$ increases by 7, when $n$ increases by 1. Therefore, the common difference $d_1$ must be 7. Similarly, $d_2=4$ .

Therefore, for $n=2$ , we have:

$\begin{aligned} \frac {2a_1+d_1}{2a_2+d_2} & = \frac {7(2)+1}{4(2)+27} \\ \frac {2a_1+7}{2a_2+4} & = \frac {15}{35} = \frac 37 \\ 14a_1 + 49 & = 6a_2+12 \\ 6a_2 - 14a_1 & = 49-12 \\ 6a_2 - 14 \times \frac 8{31} a_2 & = 37 \\ \frac {74}{31}a_2 & = 37 \\ \implies a_2 & = \frac {31}2 \\ a_1 & = 4 \end{aligned}$

Now, we have:

$\begin{aligned} X & = \frac {a_1 + (m-1)d_1}{a_2+(m-1)d_2} \\ & = \frac {4+7(m-1)}{\frac {31}2+4(m-1)} \\ & = \frac {7m-3}{4m+\frac {23}2} \\ & = \frac {14m-6}{8m+23} \end{aligned}$

$\implies X + \dfrac {6-14m}{8m+23} = \dfrac {14m-6}{8m+23} + \dfrac {6-14m}{8m+23} = \boxed{0}$

Replace n by 2n-1 in the ratio expression to get ratio of general terms

that will be 14m-6/8m+23

hence answer is zero :)