jmc

Let $a$ be the first term in this arithmetic sequence, and let $d$ be the common difference. Then the $7^{\text{th}}$ term is $a+6d=30$, and the $11^{\text{th}}$ term is $a+10d=60$. Subtracting these equations, we get $4d=30$, so $d=30/4=15/2$.

Then the $21^{\text{st}}$ term is $a+20d=(a+10d)+10d=60+10\cdot 15/2=\boxed{135}$.