The expression of the n-th term of an arithmetic sequence is
[tex]a_n=a_1+(n-1)\cdot d[/tex]The problem here is to find the initial value a1 and the value of d (common difference), then, let's write that equation for each value that we have, first, let's do n = 5
[tex]\begin{gathered} a_5=a_1+(5-1)\cdot d \\ \\ 11=a_1+4d \end{gathered}[/tex]Now let's repeat the process for n = 11
[tex]\begin{gathered} a_{11}=a_1+(11-1)\cdot d \\ \\ 29=a_1+10d \end{gathered}[/tex]Now we have a system of equations!
[tex]\begin{cases}a_1+4d=11{} \\ a_1+10d=29{}\end{cases}[/tex]To solve that system we can subtract the equations
[tex]\begin{gathered} a_1-a_1+4d-10d=11-29 \\ \\ -6d=-18 \\ \\ d=\frac{18}{6} \\ \\ d=3 \end{gathered}[/tex]We can find a1 now because we have the value of d, we can use any of the two equations we wrote to do it, I'll use the first one
[tex]\begin{gathered} a_1+4d=11 \\ \\ a_1+4\cdot3=11 \\ \\ a_1+12=11 \\ \\ a_1=-1 \end{gathered}[/tex]Now we have all the term to write the rule of the n-th term,
[tex]\begin{gathered} a_n=a_1+(n-1)\cdot d \\ \\ a_n=-1+(n-1)\cdot3 \end{gathered}[/tex]The rule for the n-th term is
[tex]a_n=-1+3\cdot(n-1)[/tex]