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: Yes, $a$ and $d$ are not integers. To be precise,$$a=-\frac{720}{41}\text{ and }d=-\frac3{41}.$$
What exactly is your question? You already know that they are fractions. Do you expect to get a different answer if you use another method?
A: Method$\#1:$
HINT:
If we put $n=13,41$ we get two simultaneous equations in $a,d$
Solve for them.
Do you how to calculate the $26$th term?
Method$\#2:$
$$S_m=\dfrac m2\{2a+(m-1)d\}$$
$$S_n=\dfrac n2\{2a+(n-1)d\}$$
$$\dfrac12\left(\dfrac{S_m}m+\dfrac{S_n}n\right)=a+\dfrac{(m+n-2)d}2$$
Now the $r$ term $$T_{r+1}=a+rd$$
$\implies 2r=m+n-2$
Here $m=13,n=41\implies r=?$
A: $a_1$ 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.
