2. The sum of the first $p$ terms of an arithmetic sequence is $q$ ($p, q$ are unequal positive integers), and the sum of the first $q$ terms is $p$. Then the sum of the first $p+q$ terms of this sequence is
A. $p+q$
B. $p-q$
C. $-p+q$
D. $-p-q$
2. Choose D.

Brief solution: $S_{n}=\frac{d}{2} n^{2}+\frac{2 a_{1}-d}{2} \cdot n$, from the given we have $\left\{\begin{array}{l}q=\frac{d}{2} p^{2}+\frac{2 a_{1}-d}{2} p \cdots \text { (1), } \\ p=\frac{d}{2} q^{2}+\frac{2 a_{1}-d}{2} q \cdots \text { (2) }\end{array}\right.$
(1) - (2) yields $q-p=\frac{d}{2}\left(p^{2}+q^{2}\right)+\frac{2 a_{1}-d}{2}(p-q)$.

Since $p-q \neq 0$, we have $-1=\frac{d}{2}(p+q)+\frac{2 a_{1}-d}{2}$.
Multiplying both sides of the above equation by $p+q$, we get $-(p+q)=\frac{d}{2}(p+q)^{2}+\frac{2 a_{1}-d}{2}(p+q)=S_{p+q}$. Thus, $S_{p-q}=-(p+q)$.