problem: The third term of an arithmetic sequence is $5$ and the sixth term is $-1.$ Find the twelfth term of this sequence.
solution: $\emph{Solution 1: Find the first term and the common difference.}$
  
  Let the first term of the sequence be $a$ and the common difference be $d.$ Then, the third term is $a+2d$ and the sixth term is $a+5d,$ so we have the system $a+2d = 5,$ $a+5d=-1.$ Subtracting the first equation from the second gives $3d = -6,$ so $d=-2.$ Substituting this into either of our original equations gives $a=9,$ so the twelfth term of the sequence is $a+11d = 9+11(-2) = \boxed{-13}.$
  
  $\emph{Solution 2: Use our understanding of arithmetic sequences.}$
  
  The sixth term is $6$ less than the third term. The twelfth term is twice as far from the sixth term $(6$ steps$)$ as the sixth term is from the third term $(3$ steps$).$ Therefore, the twelfth term is $2\cdot 6 = 12$ less than the sixth term, so it is $-1-12=\boxed{-13}.$