Step 1: Here $a = 21$ and $d = 18 - 21 = -3$.
Step 2: Use $$S_n = \frac{n}{2}\left[2a + (n-1)d\right] = 0$$
Step 3: $$\frac{n}{2}\left[2(21) + (n-1)(-3)\right] = 0$$
$$\frac{n}{2}\left[42 - 3n + 3\right] = 0 \Rightarrow \frac{n}{2}(45 - 3n) = 0$$
Step 4: Since $n \neq 0$,
$$45 - 3n = 0 \Rightarrow n = 15$$
Hence $15$ terms must be added to get the sum zero.