Given, $\left( {7n + 1} \right):\left( {4n + 27} \right)$ is the ratio of the sum of $n$ terms of two A.P.’s. Find the ratio of their ${m^{th}}$ terms.

Hint: The given question deals with arithmetic progression. An arithmetic progression or an A.P. is a sequence of numbers in which the next number can easily be obtained by adding a constant or a fixed number to the first number in order to obtain a pair of consecutive terms.

Complete step by step solution:
In order to find the sum of first $n$ terms of an A.P, the formula is $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ refers to the first term,$d$ refers to the common difference, $n$ refers to the total number of terms and $S$ is the sum of first $n$ term of an A.P.
According to the question,
We are given that $\dfrac{{{S_n}}}{{{S_n}^\prime }} = \dfrac{{7n + 1}}{{4n + 27}}$-----(1)
Let the first terms be ${a_1},{a_2}$ and similarly ${d_1},{d_2}$ be the common difference of two A.P.’s. Their sum for their $n$ terms will be given by:
${S_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]$ and
${S_n}^\prime = \dfrac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]$
So, their ratio will be:
$\dfrac{{{S_n}}}{{{S_n}^\prime }} = \dfrac{{\dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\dfrac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}}$
$\dfrac{{{S_n}}}{{{S_n}^\prime }} = \dfrac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}}$------(2)
Now, from equation (1) and (2), we get
$ \Rightarrow \dfrac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \dfrac{{7n + 1}}{{4n + 27}}$
$ \Rightarrow \dfrac{{{a_1} + \left( {\dfrac{{n - 1}}{2}} \right){d_1}}}{{{a_2} + \left( {\dfrac{{n - 1}}{2}} \right){d_2}}} = \dfrac{{7n + 1}}{{4n + 27}}$------(3)
In order to find the ratio of ${m^{th}}$ terms of two A.P.’s we thereby, replace $n$ by $\left( {2m - 1} \right)$ in equation (3):
$
  \because \dfrac{{n - 1}}{2} = m - 1 \\
   \Rightarrow n - 1 = 2m - 2 \\
   \Rightarrow n = 2m - 1 \\
 $
$
   \Rightarrow \dfrac{{{a_1} + \left( {m - 1} \right){d_1}}}{{{a_2} + \left( {m - 1} \right){d_2}}} = \dfrac{{7\left( {2m - 1} \right) + 1}}{{4\left( {2m - 1} \right) + 27}} \\
   \Rightarrow \dfrac{{{a_{m1}}}}{{{a_{m2}}}} = \dfrac{{14m - 6}}{{8m + 23}} \\
   \Rightarrow {a_{m1}}:{a_{m2}} = 14m - 6:8m + 23 \\
 $
Therefore, the ratio of the ${m^{th}}$ terms of two A.P.’s is $14m - 6:8m + 23$.

Note:
The given question was an easy one. The only trick to solve these questions is to use the formulas correctly. Students must be clear of arithmetic progression and related concepts. The formulas for finding out the sum of first$n$terms or of a finite arithmetic progression should be well known by them.