Let {T_r} be the {r^{th}} term of an A.P. for | vedclass

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $r^{th}$ term is given by ${T_r} = a + (r - 1)d$.Given,${T_m} = a + (

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$1$

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $r^{th}$ term is given by ${T_r} = a + (r - 1)d$.Given,${T_m} = a + (m - 1)d = \frac{1}{n}$  --- $(1)$Given,${T_n} = a + (n - 1)d = \frac{1}{m}$  --- $(2)$Subtracting equation $(2)$ from $(1)$:$(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$Since $m 
eq n$,we get $d = \frac{1}{mn}$.Substituting $d$ in equation $(1)$:$a + (m - 1)\frac{1}{mn} = \frac{1}{n}$$a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n}$$a = \frac{1}{mn}$.Now,the $mn^{th}$ term is ${T_{mn}} = a + (mn - 1)d$.${T_{mn}} = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = \frac{1}{mn} + 1 - \frac{1}{mn} = 1$.

Let ${T_r}$ be the ${r^{th}}$ term of an $A.P.$ for $r = 1, 2, 3, ....$. 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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