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

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(B) Let $a$ be the first term and $d$ be the common difference of the corresponding $A.P.$The $p^{th}$ term of the $A.P.$ is $T_p = a + (p - 1)d = \fr

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$1$

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(B) Let $a$ be the first term and $d$ be the common difference of the corresponding $A.P.$The $p^{th}$ term of the $A.P.$ is $T_p = a + (p - 1)d = \frac{1}{q} \dots (i)$The $q^{th}$ term of the $A.P.$ is $T_q = a + (q - 1)d = \frac{1}{p} \dots (ii)$Subtracting $(ii)$ from $(i)$:$(p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$$\Rightarrow d = \frac{1}{pq}$Substituting $d$ into $(i)$:$a + (p - 1)\frac{1}{pq} = \frac{1}{q}$$a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q}$$a = \frac{1}{pq}$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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