smc

The nth term of the sequence with first term $a$ is $a+d(n-1)$. That means the sum of the first $n$ terms is with first term $a$ is $\frac{n(2a+dn-d)}{2}$. The 2nth term of the sequence with first term $a$ is $a+d(2n-1)$. That means the sum of the first $2n$ terms is with first term $a$ is $\frac{2n(2a+2dn-d)}{2}$. The 3nth term of the sequence with first term $a$ is $a+d(3n-1)$. That means the sum of the first $3n$ terms is with first term $a$ is $\frac{3n(2a+3dn-d)}{2}$. Substituting $s_1$, $s_2$, and $s_3$ with its respective values results in $$\frac{3n(2a+3dn-d)}{2}-\frac{2n(2a+2dn-d)}{2}-\frac{n(2a+dn-d)}{2}$$ $$\frac{6an+9d{n}^2-3nd}{2}-\frac{4an+4d{n}^2-2nd}{2}-\frac{2an+d{n}^2-dn}{2}$$ $$\frac{4d{n}^2}{2}$$ $$2d{n}^2$$ The answer is $\boxed{d\text{and}n}$.