Question

[The sum of $p$ terms of an AP is $q$ and the sum of $q$ terms is $p;$ prove that the sum of $(p+q)$ terms is $-(p+q).]$

Answer 1

$\frac{p}{2}(2a+(p-1\left)d\right)=q$

$\frac{q}{2}(2a+(q-1\left)d\right)=p$

$2a+(p-1)d=2q/p---\left(i\right)$

$2a+(q-1)d=2p/q---\left(ii\right)$

$d(p-q)=\frac{2q}{p}-\frac{2p}{q}$

$d=\frac{2(q^2-p^2)}{(p-q)\left(pq\right)}=\frac{-2(p+q)}{pq}$

$2a=\frac{2q}{p}-(p-1)\times \frac{-2(p+q)}{pq}$

$\frac{2q^2+2p^2+2pq-2p-2q}{pq}$

$a=\frac{(p+q{)}^2-2(p+q)+p^2+q^2}{2pq}$

$\frac{p+q}{2}(\frac{(p+q{)}^2-2(p+q)+p^2+q^2}{2pq}+\frac{(p+q-1)\times -2(p+q)}{pq})$

$\Rightarrow \frac{p+q}{2pq}\left(\right(p+q{)}^2-2(p+q)-2p^2-2pq-2pq-2q^2+2p+2q+p^2+q^2)$

$=-(p+q)$ Proved