Step-by-step Solution:

In the equivalent Arithmetic Series, \[ \frac{1}{p} = a + (q-1)d \quad \Rightarrow \quad 1 = ap + (q-1)dp. \] Similarly, \[ \frac{1}{q} = a + (p-1)d \quad \Rightarrow \quad 1 = aq + (p-1)dq. \] Equating the two expressions: \[ ap + (q-1)dp = aq + (p-1)dq. \] Simplify: \[ a(p-q) = (p-q)d \quad \Rightarrow \quad a = d. \] Substitute \( a = d \) into the first equation: \[ \frac{1}{p} = d + (q-1)d \quad \Rightarrow \quad \frac{1}{p} = qd \quad \Rightarrow \quad d = \frac{1}{pq}. \] The \( pq \)-th term of the AP is: \[ a + (pq - 1)d = d + (pq - 1)d = pq \cdot d = 1. \]