Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Now, given mth term = 1/n And nth = 1/m a + (n-1) d = 1/m and a + (m - 1)d = 1/n By subtracting the above two equations, we get, (m-1-n+1) d = 1/n – 1/m ⇒ (m - n)d = (m-n)/mn ⇒ d = 1/mn Now, (mn)th term = a + (mn -1) d = a + (mn-1) × 1/mn = a + 1 – 1/mn ……(1) Now, a + (n-1) d = 1/m ⇒ a = 1/m – (n-1)d =1/m – (n-1) × 1/mn = 1/m – 1/m + 1/mn ∴ a = 1/mn From (1), (mn)th term = 1/mn + 1 - 1/mn = 1 Hence, option (3) is correct.