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 ½ (pq + 1) where p ≠ q.

The general term of an A.P. is given by the equation

a_n=a+(n-1)d

According to the conditions given in the question,

p^th term can be written as a_p=a+(p-1)d = \( \dfrac{1}{p}\)

q^th term can be written as a_q=a+(q-1)d = \( \dfrac{1}{q}\)

Subtract a_q from a_p 

(p-1)d-(q-1)d = \( \dfrac{1}{q}- \dfrac{1}{p}\)

(p-1-q+1)d = \( \dfrac{p-q}{pq}\)

(p-q)d = \( \dfrac{p-q}{pq}\)

\( d= \dfrac{1}{pq}\)

Substitute \( d= \dfrac{1}{pq}\)in a+(p-1)d=\( \dfrac{1}{q}\)

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

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

The sum of first n terms of an arithmetic progression is given by the equation

\( s_n= \dfrac{n}{2}[2a+(n-1)d]\)

Substitute the values of n , a and d in the equation.

\( s_{pq}= \dfrac{pq}{2}[\dfrac{2}{pq}+(pq-1)\dfrac{1}{pq}]\)

= \( 1+ \dfrac{1}{2}(pq-1)\)

= \( \dfrac{1}{2}pq+1- \dfrac{1}{2}\)

= \( \dfrac{1}{2}pq+\dfrac{1}{2}\)

= \( \dfrac{1}{2}(pq+1)\)

Therefore, \( \dfrac{1}{2}(pq+1)\) is the sum of first pq terms of the A.P.