Let the two numbers be a and b. Then,

A.M = \(\frac{a+b}{2}\)

G.M. = \(\sqrt{ab}\)

H.M. =\(\frac{2ab}{a+b}\)

Given, A.M : G.M = 5 : 4

⇒ \(\frac{\frac{(a+b)}{2}}{\sqrt{ab}}\) = \(\frac{5}{4}\) ⇒ \(\frac{a+b}{2\sqrt{ab}}\) = \(\frac{5}{4}\)                  .....(i)

Also, G.M. – H.M. = \(\frac{16}{5}\)

⇒ \(\sqrt{ab}\) - \(\frac{2ab}{a+b}\) = \(\frac{16}{5}\)                           ......(ii)

From (i), a + b = \(\frac{5\times2}{4}\)\(\sqrt{ab}\) ⇒ a + b \(\frac{5}{2}\)\(\sqrt{ab}\).                      ....(iii)

Putting this value of (a + b) in (ii), we have

\(\sqrt{ab}\) - \(\frac{2ab}{\sqrt{ab}}\) x \(\frac{2}{5}\) = \(\frac{16}{5}\) ⇒ \(\sqrt{ab}\) - \(\frac{4}{5}\)\(\sqrt{ab}\) = \(\frac{16}{5}\)

⇒ \(\frac{1}{5}\)\(\sqrt{ab}\) = \(\frac{16}{5}\) ⇒ \(\sqrt{ab}\) = 16 ⇒ ab = 256

∴ From (iii), a + b = \(\frac{5}{2}\)x 16 = 40.

∴ (a – b)² = (a + b)² – 4ab = 40² – 4 × 256 = 1600 – 1024 = 576

⇒ a – b = ± 24

Now solving a + b = 40 and a – b = ± 24, we get

a = 32, b = 8 or a = 8, b = 32. 

∴ The numbers are 8 and 32.