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In a H.P., $p^{\text {th }}$ term is $q$ and the $g^{\text {th }}$ term is $p$. Then $p q^{\text {th }}$ term is
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$1$
Let $a$ be the first term and $d$ be the common difference of the corresponding A.P.
$p^{\text {th }}$ term of A.P.$\left(T_p\right)=a+(p-1) d=\frac{1}{q}$\ldots(i)
$q^{\text {th }}$ term of A.P.$\left(T_q\right)=a+(q-1) d=\frac{1}{p}$\ldots(ii)
From (i) - (ii),$(p-q) d=\frac{1}{q}-\frac{1}{p}$ $=\frac{p-q}{p q} \Rightarrow d=\frac{1}{p q}$
From (i) $a+(p-1) \frac{1}{p q}=\frac{1}{q} \Rightarrow a=\frac{1}{p q}$
$\therefore \quad T_{p q}=a+(p q-1) d=$ $\frac{1}{p q}+(p q-1) \frac{1}{p q}=1$
Therefore $p q^{\text {th }}$ term is 1.
$p^{\text {th }}$ term of A.P.$\left(T_p\right)=a+(p-1) d=\frac{1}{q}$\ldots(i)
$q^{\text {th }}$ term of A.P.$\left(T_q\right)=a+(q-1) d=\frac{1}{p}$\ldots(ii)
From (i) - (ii),$(p-q) d=\frac{1}{q}-\frac{1}{p}$ $=\frac{p-q}{p q} \Rightarrow d=\frac{1}{p q}$
From (i) $a+(p-1) \frac{1}{p q}=\frac{1}{q} \Rightarrow a=\frac{1}{p q}$
$\therefore \quad T_{p q}=a+(p q-1) d=$ $\frac{1}{p q}+(p q-1) \frac{1}{p q}=1$
Therefore $p q^{\text {th }}$ term is 1.
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