Consecutive terms in an arithmetic sequence differ by a constant [tex]d[/tex]. So
[tex]5k-1=2k+d[/tex]
[tex]6k+2=5k-1+d[/tex]
[tex]\implies\begin{cases}3k-d=1\\k-d=-3\end{cases}\implies k=2,d=5[/tex]
Denote the [tex]n[/tex]-th term in the sequence by [tex]a_n[/tex]. Now that [tex]a_1=2[/tex], we have
[tex]a_n=a_{n-1}+5=a_{n-2}+2\cdot5=\cdots=a_1+(n-1)\cdot5[/tex]
which means
[tex]a_8=2+(8-1)\cdot5=37[/tex]
