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(A) The $r$-th term of an arithmetic progression is given by $T_r = a + (r - 1)d$.Given $T_m = 1/n$ and $T_n = 1/m$,we have:$a + (m - 1)d = 1/n$  ---

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(A) The $r$-th term of an arithmetic progression is given by $T_r = a + (r - 1)d$.Given $T_m = 1/n$ and $T_n = 1/m$,we have:$a + (m - 1)d = 1/n$  --- $(1)$$a + (n - 1)d = 1/m$  --- $(2)$Subtracting equation $(2)$ from $(1)$:$(a + (m - 1)d) - (a + (n - 1)d) = 1/n - 1/m$$(m - 1 - n + 1)d = (m - n) / (mn)$$(m - n)d = (m - n) / (mn)$Since $m 
eq n$,we can divide by $(m - n)$:$d = 1/(mn)$Now,substitute $d = 1/(mn)$ into equation $(1)$:$a + (m - 1)(1/(mn)) = 1/n$$a + 1/n - 1/(mn) = 1/n$$a = 1/(mn)$Therefore,$a - d = 1/(mn) - 1/(mn) = 0$.

The $r$-th term of an arithmetic progression is $T_r$. Its first term is $a$ and the common difference is $d$. If for some positive integers $m, n, m \neq n,$ we have $T_m = 1/n$ and $T_n = 1/m,$ then $a - d = \dots\dots.$

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