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 \[\left( \frac{pq+1}{2} \right)\].

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.