An arithmetic sequence takes the form
\(a_n=a_{n-1}+d\)
where \(d\) is the common difference between terms. You can solve for \(a_n\) in terms of any of the previous terms of the sequence:
\(a_n=a_{n-1}+d\implies~a_n=a_{n-2}+2d\implies~a_n=a_{n-3}+3d\implies\cdots\implies~a_n=a_{n-k}+kd\)
for some integer \(1\le k\le n-1\)
Continuing in this way, you know that the sequence can be defined explicitly in terms of the first term \(a_1\)
\(a_n=a_1+(n-1)d\)
Given that the 4th term is \(a_4=-6\) and the 11th term is \(a_{11}=-34\), you have the following system of equations.
\(\begin{cases}-6=a_1+(4-1)d\\-34=a_1+(11-1)d\end{cases}\)
Solving this system for the two unknowns yields \(a_1=6\) and \(d=-4\).
So, the sequence is given explicitly by
\(a_n=6+(n-1)(-4)=-4n+5\)
