What is the sum of the first $p+q $ terms of this A.P.? The sum of the first $p $ terms of an A.P. is $q $ and the sum of the first $q $ terms of the same A.P. is $p $. Find the sum of the first $p+q $ terms of the A.P.
Answer given: $$-(p+q) $$
 A: $$q=\frac{p(a+(p-1)d)}{2}\tag{1}$$
$$p=\frac{q(a+(q-1)d)}{2}\tag{2}$$
$$S_{p+q}=\frac{(p+q)(a+(p+q-1)d)}{2}$$
Do $(2)-(1)$ to get $-2=a+(p+q-1)d$
Therefore,
$$S_{p+q}=-(p+q)$$
A: A slightly unorthodox approach. 
Consider the given AP $(AP)$ with "points" $(n,S_n)$ as follows:
$$(0,0),(p,q),(q,p),(p+q,k)$$
where $k$ is an unknown to be determined. 
Transform this to another AP $(AP')$ by adding $1$ to each term of the original AP. This is the same as adding $n$ to $S_n$ of the original AP. Hence, $AP'$ has "points" $(n,S'_n)=(n,S_n+n)$ are as follows:
$$(0,0),(p,q+p),(q,p+q),(p+q,k+\overline{p+q})$$
By symmetry, as $S'_p=S'_q$, therefore $S'_{p+q}=S'_0=0$ (see note below for details). Hence $k=-(p+q)$. 
As such, for the original AP, $AP$, the sum to $(p+q)$ terms is 
$$\color{red}{S_{p+q}=k=-(p+q)}$$

Note: 
For $AP'$, $S'_p=S'_q$, which means that $S'_n$ is symmetrical about $\frac {p+q}2$ which is the "middle" term (i.e. AP' has negative common difference, terms up to/after the middle term are positive/negative respectively and the $(\frac {p+q}2\pm i)$-th term are numerically equal but of opposite sign). By symmetry, $S'_{p+q}=S'_0=0$ hence $k=-(p+q)$. 
