**Answer : **Given :if pth term of an ap is 1/q and the qth term of an ap is 1/p

To prove : the sum of pq terms is pq+1/2

Let first term and common difference be a and d respectively

Now according to the question

T_{p}=a+(p-1)d=1/q ..................eq (1)

T_{q}=a+(q-1)d=1/p ...................eq(2)

Subtracting eq 2 by eq1, we get

=> (p-q)d = (1/q )- (1/p )

=> (p-q)d = [ (p-q) / pq ]

=> d =1/pq......................eq 3

Substituting the value of d in eq 1

=> 1/q=a+[(p-1) (1/(pq)) ]

=> 1/q = a + [1/q] - [1/(pq)]

=> a=1/pq.................................eq 4

Now the sum of pq terms using the formula S_{n} = n/2 ( 2a +(n-1) d ) is given by

=> S_{pq}=(pq/2 ) [ (2/(pq)) + (pq-1) (1/pq)] {using the value of eq 3 and eq 4}

=> S_{pq}= (pq/2 ) [ (2/(pq)) + 1 -(1/pq)]

=> S_{pq}= (pq/2 ) [ (1/(pq)) + 1]

=> S_{pq }= ( 1/2 ) + (pq /2)

**=> S**_{pq}=(pq+1)/2