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Question Number 21502 by 786786AM last updated on 25/Sep/17

$$\mathrm{In}\mathrm{an}\mathrm{A}.\mathrm{P},\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{p}\mathrm{terms}\mathrm{is}\mathrm{q}\mathrm{and}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{q}\mathrm{terms}\mathrm{is}\mathrm{p}.$$$$\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}(\mathrm{p}+\mathrm{q})\mathrm{terms}\mathrm{is}-(\mathrm{p}+\mathrm{q}).$$

Answered by $@ty@m last updated on 25/Sep/17

$$ATQ$$$${\mathbf{S}}_p=q$$$$\frac{p}{2}[2a+(p-1)d]=q$$$$\Rightarrow 2a+(p-1)d=\frac{2q}{p}--\left(1\right)$$$${\mathbf{S}}_q=p$$$$\frac{q}{2}[2a+(q-1)d]=p$$$$\Rightarrow 2a+(q-1)d=\frac{2p}{q}--\left(2\right)$$$$Subtracting\left(2\right)from\left(1\right)$$$$(p-1-q+1)d=\frac{2q}{p}-\frac{2p}{q}$$$$(p-q)d=\frac{2(q^2-p^2)}{pq}$$$$d=\frac{-2(p+q)}{pq}--\left(3\right)$$$$\therefore from\left(1\right),$$$$2a+(p-1)\frac{-2(p+q)}{pq}=\frac{2q}{p}$$$$2a=\frac{2(p-1)(p+q)}{pq}+\frac{2q}{p}---\left(4\right)$$$$\therefore {\mathbf{S}}_{p+q}=\frac{p+q}{2}[2a+(p+q-1)d]$$$$\Rightarrow {\mathbf{S}}_{p+q}=\frac{p+q}{2}[\frac{2(p-1)(p+q)}{pq}+\frac{2q}{p}+(p+q-1)\frac{-2(p+q)}{pq}]$$$$from\left(3\right)\mathrm{\&}\left(4\right)$$$$=\frac{2(p+q)}{2pq}[p^2-p+pq-q+q^2-{(p+q)}^2+p+q]$$$$=\frac{p+q}{pq}\times (-pq)$$$$=-(p+q)$$$$\mathrm{Q}.\mathrm{E}.\mathrm{D}.$$

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