efficient way to solve such questions in arithmetic progression?

Question

QUESTION: The sum of $p$ terms of an A.P is $q$ ,and the sum of $q$ terms is $p$.Find the sum of $p+q$ terms
MY ATTEMPT:I know it is possible to solve this question by using the formula for the sum of n terms of an A.P and substitute q from equation 1 below into equation 2 and solve for p and q.This is a very long process and i was wondering if there is any easier way to solve these sort of problems. Any help is appreciated.Thanks :) $$ \frac{p}{2} (2a+(p-1)d)=q $$ equation 1 $$ \frac{q}{2} (2a+(q-1)d)=p $$ equation 2

Answer 1

\begin{align} \frac p2 (2a+(p-1)d) &= q \\ \frac q2 (2a+(q-1)d) &= p \\ \hline 2a+(p-1)d &= 2\frac qp \\ 2a+(q-1)d &= 2\frac pq \\ \hline (p-q)d &= 2\frac qp - 2\frac pq \\ d &= 2\frac{q^2-p^2}{(p-q)pq} \\ d &= -2\frac{p+q}{pq} \\ \end{align}

If you don't require that $d$ be an integer, then there are an infinite number of such sequences.

If you want $d$ to be an integer. Then your choices are pretty limited.

Answer 2

$$\begin{align} S_n=\ &An^2&&+Bn &&&&\scriptsize \left(A=\frac d2; \qquad B=a-\frac d2\right)\\\\ S_p=\ &Ap^2&&+Bp&&=q\tag{1}\\ S_q=\ &Aq^2&&+Bq&&=p\tag{2}\\\\ (1)-(2):\qquad &\ A(p^2-q^2)&&+B(p-q)&&=q-p &&\scriptsize (p\neq q)\\ &\ A(p+q)&&+B&&=-1\\\\ S_{p+q}=\ &A(p+q)^2 &&+B(p+q) &&=\color{red}{-(p+q)} \end{align}$$