Let the original numbers be

a, a + d, and a + 2d

According to the question –

a + a + d + a + 2d = 15

⇒ 3a + 3d = 15 or a + d = 5

⇒ d = 5 – a

After the addition, the three numbers are:

a + 1, a + d + 3, and a + 2d + 9

they are now in GP, that is –

⇒

⇒ (a + d + 3)^{2 =} (a + 2d + 9)(a + 1)

⇒ a² + d² + 9 + 2ad + 6d + 6a = a² + a + 2da + 2d + 9a + 9

⇒ (5 – a)² – 4a + 4(5 – a) = 0

⇒ 25 + a² – 10a – 4a + 20 – 4a = 0

⇒ a² – 18a + 45 = 0

⇒ a² – 15a – 3a + 45 = 0

⇒ a(a – 15) – 3(a – 15) = 0

⇒ a = 3 or a = 15

∴ d = 5 – a

d = 5 – 3 or d = 5 – 15

d = 2 or – 10

∴ The numbers are 3,5,7 or 15,5, – 5