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In an A.P., if p th term is 1 q and q th term is 1 p , prove that the sum of first p q terms is 1 by 2 (p q +1), where p is not equal to q.

5. In an A.P., if pth term is 1/q  and qth term is 1/p  , prove that the sum of first pq terms is 1/2  (pq +1), where p \neq q

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Given : In an A.P., if pth term is 1/q  and qth term is 1/p 

a_p=a+(p-1)d=\frac{1}{q}.................(1)

a_q=a+(q-1)d=\frac{1}{p}.................(2)

Subtracting (2) from (1), we get

           \Rightarrow \, \, a_p-a_q

 \Rightarrow \, \, (p-1)d-(q-1)d=\frac{1}{q}-\frac{1}{p}

\Rightarrow \, \, pd-d-qd+d=\frac{p-q}{pq}

\Rightarrow \, \, (p-q)d=\frac{p-q}{pq}

\Rightarrow \, \, d=\frac{1}{pq}

 Putting value of d in equation (1),we get 

a+(p-1)\frac{1}{pq} = \frac{1}{q}

\Rightarrow a+\frac{1}{q}-\frac{1}{pq} = \frac{1}{q}

\Rightarrow a= \frac{1}{pq}

\therefore \, \, S_p_q=\frac{pq}{2}[2.\frac{1}{pq}+(pq-1).\frac{1}{pq}]

\Rightarrow \, \, S_p_q=\frac{1}{2}[2+(pq-1)]

\Rightarrow \, \, S_p_q=\frac{1}{2}[pq+1]

Hence,the sum of first pq  terms is 1/2  (pq +1), where p \neq q.

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