Let a is the first term and d is the common difference of the AP

Given, pth term of AP is q

=> a + (p - 1)d = q    .................1

and qth term of AP is p

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

Subtracting equation 2 from 1, we get

=> a + (p - 1)d - {a + (q - 1)d} = q - p

=> a + pd - d - a - qd + d = q - p

=> pd - qd = q - p

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

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

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

=> d = -1

Now add equation 1 and 2, we get

=> 2a + (p + q - 2)d = p + q

=> 2a + (p + q -2) * (-1) = p + q          {since d = -1}

=> 2a - p - q + 2 = p + q

=> 2a = p + q - 2 + p + q

=> 2a = 2(p + q - 1)

=> a = p + q - 1

=> a - (p + q - 1) = 0

=> a + (p + q - 1)*(-1) = 0                   

=> a + (p + q - 1)d = 0           {since d = -1}

which is the (p + q)th term of the AP and is equal to 0