1. The answer is $(\mathbf{D})$.

It must be $a+b=0$; substituting yields $\frac{c}{2}-\frac{c}{3}=0$, that is $\frac{c}{6}=0$. The only possibility is that $c=0$.
1) The sum of two numbers $a$ and $b$ is equal to zero. Knowing that there exists a number $c$ such that $a=\frac{c}{2}$ and $b=-\frac{c}{3}$, determine the value of $a$.
(A) -6 ,
(B) -3 ,
(C) -2 ,
(D) 0, (E) 2 .