Answer : (c) 0
Given,
Tm = \(\frac{1}{n}\) ⇒ a + (m-1)d = \(\frac{1}{n}\) ...(i)
Tn = \(\frac{1}{m}\) ⇒ a + (n-1)d = \(\frac{1}{m}\) ... (ii)
Eq. (ii) – Eq. (i)
⇒ (n – m)d = \(\frac{1}{m}\) - \(\frac{1}{n}\)
⇒ (n - m)d = \(\frac{n-m}{mn}\)
⇒ d = \(\frac{1}{mn}\)
Putting d = \(\frac{1}{mn}\) in (i), we get a = \(\frac{1}{mn}\)
∴ a – d = \(\frac{1}{mn}\) - \(\frac{1}{mn}\) = 0