<s>[INST] The ratio of their $n$-th term. The sum of $n$ terms of two arithmetic series are in the ratio of $(7n+ 1) : (4n+ 27)$. We have to find the ratio of their $n$-th term.
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I tried to find the ratio by using the formula of summation of A.P.
But it becomes too long due to many variables that is $a_1,a_2,d_1,d_2$
 [/INST] Let $a_1$, $a_2$ be the first terms and $d_1$, $d_2$ the common differences of the given APs. Then, their sum of $n$ terms are given by: $S_{n}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]$ and $S_{n}^{\prime}=\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]$
On dividing, we get:
$$
\frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{(7 n+1)}{(4 n+27)}
$$
$\therefore$ The ratio of their 11 th terms $=\frac{a_{1}+10 d_{1}}{a_{2}+10 d_{2}}=\frac{2 a_{1}+20 d_{1}}{2 a_{2}+20 d_{2}}$
$=\frac{2 a_{1}+(21-1) d_{1}}{2 a_{2}+(21-1) d_{2}}=\frac{(7 \times 21+1)}{(4 \times 21+27)}=\frac{148}{111} \quad[$ Putting $n=21$ in (iii) ]
Hence, the required ratio is $148: 111$.
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