Let a be the first term and d be the common difference of the given AP. Then,
`T_(m) = a + (m-1)d "and" T_(n) = a + (n-1)d.`
Now, `T_(m) = (1)/(n) "and" T_(n) = (1)/(m)` (given).
`therefore a + (m -1)d = (1)/(n) " "...(i)`
`"and" a+ (n-1)d = (1)/(m) " "... (ii)`
On subtracting (ii) form (i), we get
`(m-n)d = ((1)/(n) - (1)/(m)) = ((m-n)/(mn)) rArr d = (1)/(mn).`
Putting `d = (1)/(mn)` in (i), we get
`a + ((m-1))/(mn) = (1)/(n) rArr a = {(1)/(n) - ((m-1))/(mn)} = (1)/(mn).`
Thus, `a = (1)/(mn) "and" d = (1)/(mn).`
`therefore (mn)"th term" = a+ (mn-1)d`
` = {(1)/(mn) + ((mn-1))/(mn)} [because a = (1)/(mn)]`
= 1.
Hence, the (mn)th term of the given AP is 1.
