The ratio of the sums of first $n$ terms of two $A.P.s$ is $(7n + 1) : (4n + 27)$. Find the ratio of their $m^{th}$ terms.

(D) Let the first terms of the two $A.P.s$ be $a_1$ and $a_2$,and their common differences be $d_1$ and $d_2$ respectively.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.
Given,$\frac{S_{n1}}{S_{n2}} = \frac{\frac{n}{2} [2a_1 + (n - 1)d_1]}{\frac{n}{2} [2a_2 + (n - 1)d_2]} = \frac{7n + 1}{4n + 27}$.
This simplifies to $\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{7n + 1}{4n + 27}$.
We need to find the ratio of the $m^{th}$ terms: $\frac{a_m}{a'_m} = \frac{a_1 + (m - 1)d_1}{a_2 + (m - 1)d_2}$.
Multiply the numerator and denominator of the required ratio by $2$: $\frac{2a_1 + 2(m - 1)d_1}{2a_2 + 2(m - 1)d_2}$.
Comparing this with the given sum ratio,we set $n - 1 = 2(m - 1)$,which gives $n = 2m - 1$.
Substitute $n = 2m - 1$ into the ratio $\frac{7n + 1}{4n + 27}$:
Ratio $= \frac{7(2m - 1) + 1}{4(2m - 1) + 27} = \frac{14m - 7 + 1}{8m - 4 + 27} = \frac{14m - 6}{8m + 23}$.
Thus,the ratio of the $m^{th}$ terms is $(14m - 6) : (8m + 23)$.