problem: In an arithmetic sequence, the 7th term is 30, and the 11th term is 60. What is the 21st term of this sequence?
solution: 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 $a + 20d = (a + 10d) + 10d = 60 + 10 \cdot 15/2 = \boxed{135}$.