Let the three numbers be .

∴ According to the question

⇒ …(1)

⇒ …(2)

From 2 we get

⇒ a³ = – 1

∴ a = – 1.

From 1 we get

⇒

⇒ 12a + 12ar + 12ar² = 13r …(3)

Substituting a = – 1 in 3 we get

⇒ 12( – 1) + 12( – 1)r + 12( – 1)r² = 13r

⇒12r² + 25r + 12 = 0

⇒ 12r² + 16r + 9r + 12 = 0…(4)

⇒ 4r(3r + 4) + 3(3r + 4) = 0

∴ r = or r =

∴ Now the equation will be

⇒

⇒

∴ The three numbers are .