• # question_answer If the ${{p}^{th}}$ term of an A.P. is $\frac{1}{q}$ and ${{q}^{th}}$ term is $\frac{1}{p}$, prove that the sum of first $pq$ terms if the A.P. is $\left( \frac{pq+1}{2} \right)$.

 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.