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

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(C) Given that the $m^{th}$ term of a $H.P.$ is $n$ and the $n^{th}$ term is $m$.For the corresponding $A.P.$,the $m^{th}$ term is $\frac{1}{n}$ and t

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$\frac{mn}{r}$

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(C) Given that the $m^{th}$ term of a $H.P.$ is $n$ and the $n^{th}$ term is $m$.For the corresponding $A.P.$,the $m^{th}$ term is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$.Let $a$ be the first term and $d$ be the common difference of this $A.P.$Then,$a + (m - 1)d = \frac{1}{n}$  --- $(i)$And,$a + (n - 1)d = \frac{1}{m}$  --- (ii)Subtracting equation (ii) from $(i)$:$(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$Thus,$d = \frac{1}{mn}$.Substituting $d$ in equation $(i)$:$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}$.The $r^{th}$ term of the $A.P.$ 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 $H.P.$ is the reciprocal of the $r^{th}$ term of the $A.P.$,which is $\frac{mn}{r}$.

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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