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