Answer:
Let a be first term and d is common difference Then \[{{a}_{p}}=\frac{1}{q}\Rightarrow a+(p-1)d=\frac{1}{q}\] ?.(i) \[{{a}_{q}}=\frac{1}{p}\Rightarrow a+(q-1)d=\frac{1}{p}\] ?(ii) Subtracting eq. (ii) from eq. (i) \[pd-qd+=\frac{1}{q}-\frac{1}{p}=\frac{p-q}{pq}\] \[(p-q)d=\frac{p-q}{pq}\] or \[d=\frac{1}{pq}\] Putting value of d in eq. (i) \[a+(p-1)\frac{1}{pq}=\frac{1}{q}\Rightarrow a=\frac{1}{q}-\frac{p}{pq}+\frac{1}{pq}\] \[a=\frac{1}{pq}\] Now, \[{{S}_{pq}}=\frac{pq}{2}(2a+(pq-1)d)\] \[=\frac{pq}{2}\left( \frac{2}{pq}+(pq-1)\frac{1}{pq} \right)\] \[=\frac{pq}{2}\left( \frac{2}{pq}+\frac{pq}{pq}-\frac{1}{pq} \right)\] \[{{S}_{pq}}=\frac{pq}{2}\left( \frac{1+pq}{pq} \right)\] \[=\frac{(pq+1)}{2}\] Hence Proved.
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