Proof:

Let's consider an arithmetic progression (A.P) with the first term as 'a' and the common difference as 'd'. The general formula for the nth term of an A.P is given by:

a_n = a + (n-1)d ...(1)

Given that the pth term of the A.P is 1/q, we can substitute the values in equation (1):

a_p = a + (p-1)d = 1/q ...(2)

Similarly, substituting the values for the qth term:

a_q = a + (q-1)d = 1/p ...(3)

Now, we need to find the sum of the pq terms of the A.P, which can be represented as:

S_pq = a_p + a_p+d + a_p+2d + ... + a_p+(q-1)d

Step 1: Finding the common difference:
Subtracting equation (2) from equation (3):

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

Simplifying the equation:

d(q-p) = (q-p)/(pq)

Since (q-p) is not equal to zero (as given), we can cancel it from both sides of the equation:

d = 1/(pq)

Step 2: Finding the first term:
Substituting the value of d in equation (2):

a + (p-1)/(pq) = 1/q

Simplifying the equation:

a = (1/q) - (p-1)/(pq)

Step 3: Evaluating the sum:
Substituting the values of a and d in the sum formula:

S_pq = [(1/q) - (p-1)/(pq)] + [(1/q) - (p-1)/(pq) + 1/(pq)] + [(1/q) - (p-1)/(pq) + 2/(pq)] + ...

Simplifying the sum:
By factoring out (1/q) from each term, we get:

S_pq = (1/q)[1 + 1 + 2 + 3 + ... + (q-1)] - [(p-1)/(pq) + 2(p-1)/(pq) + 3(p-1)/(pq) + ...]

The sum of the first (q-1) natural numbers can be represented as:

1 + 2 + 3 + ... + (q-1) = (q-1)q/2

Substituting this value:

S_pq = (1/q)[(q-1)q/2] - [(p-1)/(pq) + 2(p-1)/(pq) + 3(p-1)/(pq) + ...]

Simplifying further by factoring out