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Solution

Key Concept: If a sequence is in HP, then the reciprocals of its terms are in AP.

Step 1: Convert HP to AP
Let the corresponding AP have first term $a$ and common difference $d$.

$p^{th}$ term of HP $= q$
$\Rightarrow p^{th}$ term of AP $= \dfrac{1}{q}$

$\Rightarrow a + (p-1)d = \dfrac{1}{q} \quad \cdots (1)$

$q^{th}$ term of HP $= p$
$\Rightarrow q^{th}$ term of AP $= \dfrac{1}{p}$

$\Rightarrow a + (q-1)d = \dfrac{1}{p} \quad \cdots (2)$

Step 2: Subtract equation (1) from (2)
$(q - p)d = \dfrac{1}{p} - \dfrac{1}{q}$

$(q - p)d = \dfrac{q - p}{pq}$

$d = \dfrac{1}{pq}$

Step 3: Find $a$ using equation (1)
$a + (p-1) \times \dfrac{1}{pq} = \dfrac{1}{q}$

$a = \dfrac{1}{q} - \dfrac{p-1}{pq}$

$a = \dfrac{p - (p-1)}{pq}$

$a = \dfrac{1}{pq}$

Step 4: Find the $pq^{th}$ term of AP
$T_{pq} = a + (pq - 1)d$

$= \dfrac{1}{pq} + (pq-1) \times \dfrac{1}{pq}$

$= \dfrac{1}{pq} + \dfrac{pq - 1}{pq}$

$= \dfrac{1 + pq - 1}{pq}$

$= \dfrac{pq}{pq} = 1$

Step 5: Convert back to HP
$pq^{th}$ term of HP $= \dfrac{1}{T_{pq}} = \dfrac{1}{1} = 1$

Answer: The $pq^{th}$ term of the HP is $\boxed{1}$