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(C) Let the first term be $a$ and the common difference be $d$.Given $T_m = a + (m - 1)d = \frac{1}{n}$ and $T_n = a + (n - 1)d = \frac{1}{m}$.Subtrac

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$1$

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(C) Let the first term be $a$ and the common difference be $d$.Given $T_m = a + (m - 1)d = \frac{1}{n}$ and $T_n = a + (n - 1)d = \frac{1}{m}$.Subtracting the two equations: $(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$.Thus,$d = \frac{1}{mn}$.Substituting $d$ into the first equation: $a + (m - 1)\frac{1}{mn} = \frac{1}{n} \implies a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n} \implies a = \frac{1}{mn}$.Now,$T_{mn} = a + (mn - 1)d = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1$.

Let $T_r$ be the $r^{th}$ term of an $A.P.$ for $r = 1, 2, 3, \dots$. If for some positive integers $m, n$ we have $T_m = \frac{1}{n}$ and $T_n = \frac{1}{m}$,then $T_{mn}$ equals

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