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(C) Let the first term be $a$ and the common difference be $d$.Given that the $m^{th}$ term is $a + (m - 1)d = 1/n$ (Equation $1$).Given that the $n^{

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$\frac{1}{2}(mn + 1)$

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(C) Let the first term be $a$ and the common difference be $d$.Given that the $m^{th}$ term is $a + (m - 1)d = 1/n$ (Equation $1$).Given that the $n^{th}$ term is $a + (n - 1)d = 1/m$ (Equation $2$).Subtracting Equation $2$ from Equation $1$ gives $(m - n)d = 1/n - 1/m = (m - n)/mn$,so $d = 1/(mn)$.Substituting $d$ into Equation $1$: $a + (m - 1)(1/mn) = 1/n$ $\Rightarrow a + 1/n - 1/(mn) = 1/n$ $\Rightarrow a = 1/(mn)$.The sum of the first $mn$ terms is $S_{mn} = \frac{mn}{2} [2a + (mn - 1)d]$.Substituting $a = 1/(mn)$ and $d = 1/(mn)$:$S_{mn} = \frac{mn}{2} [2(1/mn) + (mn - 1)(1/mn)] = \frac{mn}{2} [\frac{2 + mn - 1}{mn}] = \frac{mn}{2} [\frac{mn + 1}{mn}] = \frac{1}{2}(mn + 1)$.

If the $m^{th}$ term of an arithmetic progression is $1/n$ and the $n^{th}$ term is $1/m$,then the sum of the first $mn$ terms is:

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