The larger of two consecutive odd integers is three times the smaller. What is their sum?
$\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20$
Let $n$ be the smaller term. Then $n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1$

Thus, the answer is $1+(1+2)=4 \mathrm{(A)}$