If the m^{th} term of a H.P. is n and the n^{ | vedclass

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(C) Given $T_m = n$ and $T_n = m$ for a $H.P.$Therefore,for the corresponding $A.P.$,the $m^{th}$ term is $\frac{1}{n}$ and the $n^{th}$ term is $\fra

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$\frac{mn}{r}$

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(C) Given $T_m = n$ and $T_n = m$ for a $H.P.$Therefore,for the corresponding $A.P.$,the $m^{th}$ term is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$.Let $a$ and $d$ be the first term and common difference of this $A.P.$,then:$a + (m - 1)d = \frac{1}{n}$  ---$(i)$$a + (n - 1)d = \frac{1}{m}$  ---(ii)Subtracting (ii) from $(i)$:$(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$$d = \frac{1}{mn}$Substituting $d$ in $(i)$:$a + (m - 1)\frac{1}{mn} = \frac{1}{n}$$a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - m + 1}{mn} = \frac{1}{mn}$Now,the $r^{th}$ term of the corresponding $A.P.$ is:$T_r = a + (r - 1)d = \frac{1}{mn} + (r - 1)\frac{1}{mn} = \frac{1 + r - 1}{mn} = \frac{r}{mn}$Therefore,the $r^{th}$ term of the corresponding $H.P.$ is the reciprocal of the $A.P.$ term:$T_r(H.P.) = \frac{mn}{r}$.

If the $m^{th}$ term of a $H.P.$ is $n$ and the $n^{th}$ term is $m$,then the $r^{th}$ term will be

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