Question

If the mth term of an H.P. is n and nth term is m, show that the rth term is \(\frac{mn}{r}.\)

Answer 1

Let the corresponding A.P. be a, a + d, a + 2d, .... 

Since the mth term and nth term of the H.P. are n and m respectively, then for the A.P., 

mth term = a + (m – 1) d = \(\frac{1}{n}\)

nth term = a + (n – 1) d = \(\frac{1}{m}\)

(ii) – (i) ⇒ (n – 1)d – (m – 1)d = \(\frac{1}{m}\) - \(\frac{1}{n}\)

⇒ (n – m)d = \(\frac{n-m}{mn}\) ⇒ d = \(\frac{1}{mn}\)

Putting in (i), a + (m - 1) x \(\frac{1}{mn}\) = \(\frac{1}{n}\)

⇒ a = \(\frac{1}{n}\) - \(\frac{1}{n}\) + \(\frac{1}{mn}\) = \(\frac{1}{mn}\)

∴ t_r = a + (r - 1)d = \(\frac{1}{mn}\)+ (r - 1) \(\frac{1}{mn}\) = \(\frac{r}{mn}\)

∴ rth term of H.P. = \(\frac{mn}{r}\).