In a H.P.,the p^{th} term is q and the q^{th} | vedclass

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(B) Let the first term of the corresponding $A.P.$ be $A$ and the common difference be $D$. Since the terms of $H.P.$ are the reciprocals of the terms

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$1$

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(B) Let the first term of the corresponding $A.P.$ be $A$ and the common difference be $D$. Since the terms of $H.P.$ are the reciprocals of the terms of $A.P.$,we have:$T_p = \frac{1}{q} \implies A + (p - 1)D = \frac{1}{q} \quad (i)$$T_q = \frac{1}{p} \implies A + (q - 1)D = \frac{1}{p} \quad (ii)$Subtracting $(ii)$ from $(i)$:$(p - q)D = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$$D = \frac{1}{pq}$Substituting $D$ in $(i)$:$A + (p - 1)\frac{1}{pq} = \frac{1}{q}$$A = \frac{1}{q} - \frac{p - 1}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq}$Now,the $(pq)^{th}$ term of the $A.P.$ is:$T_{pq} = A + (pq - 1)D = \frac{1}{pq} + (pq - 1)\frac{1}{pq} = \frac{1 + pq - 1}{pq} = 1$Since the $(pq)^{th}$ term of the $A.P.$ is $1$,the $(pq)^{th}$ term of the $H.P.$ is the reciprocal of $1$,which is $1$.

In a $H.P.$,the $p^{th}$ term is $q$ and the $q^{th}$ term is $p$. Then the $(pq)^{th}$ term is

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