If ${\mathrm{p}}^{\mathrm{th}}$ term of an A.P. is $\frac{1}{\mathrm{q}}$ and ${\mathrm{q}}^{\mathrm{th}}$ term of A.P. is $\frac{1}{\mathrm{p}}$ then sum of the first pq terms is 
 

$$\begin{gathered} \mathrm{a}+(\mathrm{p}-1)\mathrm{d}=\frac{1}{\mathrm{q}} -\left(1\right) \\ \mathrm{a}+(\mathrm{q}-1)=\frac{1}{\mathrm{p}} -\left(2\right) \\ \left(1\right)-\left(2\right) \Rightarrow (\mathrm{p}-1-\mathrm{q}+1) \mathrm{d}=\frac{1}{\mathrm{q}}-\frac{1}{\mathrm{p}} \Rightarrow \mathrm{d}=\frac{1}{\mathrm{pq}} \\ \mathrm{a}=\frac{1}{2}-(\mathrm{p}-1)\mathrm{d}=\frac{1}{\mathrm{q}}-\frac{\mathrm{p}-1}{\mathrm{pq}}=\frac{\mathrm{p}-\mathrm{p}+1}{\mathrm{pq}}=\frac{1}{\mathrm{pq}} \\ \mathrm{Spq}=\frac{\mathrm{pq}}{2}\left(2\mathrm{a}+(\mathrm{pq}-1)\mathrm{d}\right)=\frac{\mathrm{pq}}{2}\left[\frac{2}{\mathrm{pq}}+\frac{\mathrm{pq}-1}{\mathrm{pq}}\right] \\ =\frac{1}{2}\left(\mathrm{pq}+1\right) \end{gathered}$$