Solution:

Given pth term = 1/q
That is ap = a + (p - 1)d = 1/q
aq + (pq - q)d = 1  --- (1)
Similarly, we get ap + (pq - p)d = 1  --- (2)
From (1) and (2), we get
aq + (pq - q)d = ap + (pq - p)d 
aq - ap = d[pq - p - pq + q]
a(q - p) = d(q - p)
Therefore, a = d
Equation (1) becomes,
dq + pqd - dq = 1  
d = 1/pq
Hence a = 1/pq
Consider, Spq = (pq/2)[2a + (pq - 1)d]
                    = (pq/2)[2(1/pq) + (pq - 1)(1/pq)]
                    = (1/2)[2 + pq - 1]
                    = (1/2)[pq + 1]

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