An arithmetic sequence “The sum of the first $13$ terms of an arithmetic sequence is $-234$ and the sum of the first $41$ terms is $-780$. Find the $27^\text{th}$ term of the sequence.”
The $S_n=\frac n2(2a+(n-1)d)$ formula ($S_n$ represents the sum of the first to $n^\text{th}$ terms) only gives me fractional values for $a$ (which is the first term) and $d$, which is the common difference. Are there other viable methods?

$a_1$ and $d$ are unknown, but we know:
$\frac{13}{2}(a_1 + a_{13}) = -234 \implies a_1 + a_{13} = - \frac{234\times 2}{13} = -36$
and also $\frac{41}{2}(a_1 + a_{41}) = -780 \implies a_1 + a_{41} = -38\frac{2}{41}$
and also $a_{13}= a_1 + 12d$
and $a_{41} = a_1 + 40d$.
So we get two linear equations in two variables, by  substiting the last two into the first two. There is no problem with fractions, or even arbitrary reals being used as $a_1$ or $d$. That some sums of fractions are integers is OK, it doesn't mean all solutions are integers.
