Question

If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be
(a) 0
(b) p − q
(c) p + q
(d) − (p + q)

Answer 1

(d) $-\left(p+q\right)$

$$\begin{gathered} S_p=q \\ \Rightarrow \frac{p}{2}\left\{2a+\left(p-1\right)d\right\}=q \\ \Rightarrow 2ap+\left(p-1\right)pd=2q.....\left(1\right) \\ {S}_q=p \\ \Rightarrow \frac{q}{2}\left\{2a+\left(q-1\right)d\right\}=p \\ \Rightarrow 2aq+\left(q-1\right)qd=2p.....\left(2\right) \\ \mathrm{Multiplying}\mathrm{equation}\left(1\right)\mathrm{by}q\mathrm{and}\mathrm{equation}\left(2\right)\mathrm{by}p\mathrm{and}\mathrm{then}\mathrm{solving},\mathrm{we}\mathrm{get}: \\ d=\frac{-2\left(p+q\right)}{pq} \\ \mathrm{Now},S_{p+q}=\frac{\left(p+q\right)}{2}\left[2a+\left(p+q-1\right)d\right] \\ =\frac{p}{2}\left[2a+\left(p-1\right)d+qd\right]+\frac{q}{2}\left[2a+\left(q-1\right)d+pd\right] \\ =S_p+\frac{pqd}{2}+S_q+\frac{pqd}{2} \\ =p+q+pqd \\ =p+q-\frac{2\left(p+q\right)pq}{pq} \\ =-(p+q) \end{gathered}$$