Let's assume an arithmetic progression with first term=a and common difference r: a, a+r, a+2r, etc.
The pth term is a+(p-1)r and the qth term is a+(q-1)r. So a+(p-1)r=q and a+(q-1)r=p. Subtracting one equation from the other we have (p-q)r=q-p; so q=p (trivial case), or r=(q-p)/(p-q)=-1. Also a-(p-1)=q, so a-p+1=q and a=p+q-1. The AP is p+q-1, p+q-2, p+q-3, ..., p+q-n. The nth term is therefore p+q-n.