The sum of three

Let the original numbers be

a, a + d, and a + 2d

According to the question –

3a + 3d = 21 or a + d = 7.

⇒ d = 7 – a

After the addition, the three numbers are:

a, a + d – 1, and a + 2d + 1

they are now in GP, that is –

⇒

⇒ (a + d – 1)^{2 =} a(a + 2d + 1)

⇒ a² + d² + 1 + 2ad – 2d – 2a = a² + a + 2da

⇒ (7 – a)² – 3a + 1 – 2(7 – a) = 0

⇒ 49 + a² – 14a – 3a + 1 – 14 + 2a = 0

⇒ a² – 15a + 36 = 0

⇒ a² – 12a – 3a + 36 = 0

⇒ a(a – 12) – 3(a – 12) = 0

⇒ a = 3 or a = 12

∴ d = 7 – a

d = 7 – 3 or d = 7 – 12

d = 4 or – 5

∴ The numbers are 3,7,11 or 12,7,2