I have an arithmetic progression. $$a = -27, -19, -11, ...$$ The question asks me to find a term in the progression where the term is equal to the sum of the terms behind him. Therefore: I need to find $n$ for $$a_n = S_{n-1}$$
$$a_1 + d(n-1) = \frac{n-1}{2}(2a_1 + d(n-2)) $$ $$-27 -8(n-1) = \frac{n-1}{2}(-54 -8(n-2))$$ With this I don't get to neither answer that could be logic. What are my mistakes?
Note that $$S_{n-1}=\frac{n-1}{2}\left(2a_1+(n-2)d\right).$$
Note that $d=\color{red}{+}8$.
$$a_n=S_{n-1}$$ $$a_1+d(n-1)=\frac{n-1}{2}(2a_1+d(n-2))$$ $$-27+8(n-1)=\frac{n-1}{2}(-54+8(n-2))$$ $$2(-27+8n-8)=(n-1)(-54+8n-16)$$ $$4n^2-47n+70=0$$ $$(4n-7)(n-10)=0$$ $$n=10$$