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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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