If $p, q$ are distinct natural numbers, and in an arithmetic sequence {${a_n}$}, $S_p = S_q$, where $S_k$ denotes the sum of the first $k$ terms of the sequence {${a_n}$}. Prove that $S_{p+q}=0$.
The formula for an arithmetic series is $S_n = n a_1 + \frac{n(n-1)}{2}d$. So, $S_p = p a_1+ \frac{p(p-1)}{2}d$ and $S_q = q a_1+ \frac{q(q-1)}{2}d$. Since $S_p = S_q$, then $(p – q)a_1 + \frac{(p (p – 1) – q(q-1))d}{2}=0$. This rearranges to $(p-q) a_1+\frac{((p-q)(p+q-1))d}{2}$. Since $p \neq q$, we can divide by $p-q$ to get $a_1 + \frac{(p+q-1)d}{2} = 0$. Now we take $S_{p+q}$, which by the formula turns out to be $(p+q) a_1+ \frac{(p+q)(p+q-1)d}{2}$. Factor out the $p+q$ to get $S_{p+q}=(p+q)(a_1+\frac{(p+q-1)d}{2})$. However, $a_1+\frac{(p+q-1)d}{2}=0$, so $S_{p+q}=(p+q)(0)=0$.