Let the first term be a – d

the second term be a

the third term be a + d

Given sum of consecutive three term is 27

⇒ (a – d) + a + (a + d) = 27

⇒ 3 a = 27

⇒ a = 27/ 3 = 9

Also, given product of three consecutive term is 504

⇒ (a – d) × a × (a + d) = 504

⇒ (9 – d) × 9 × (9 + d) = 504 (since, a = 9)

⇒ (9 – d) × (9 + d) = 504/ 9 = 56

⇒ 9^{2} – d^{2} = 56 (since, (a – b) (a + b) = a^{2} – b^{2})

⇒ 81 – d^{2} = 56

⇒ d^{2} = 81 – 56 = 25

⇒ d = √25 = ± 5

Case 1:

Thus, if a = 9 and d = 5

Then the three terms are,

First term a – d = 9 – 5 = 4

Second term a = 9

Third term a + d = 9 + 5 = 14

Thus, the A.P. is 4, 9, 14

Case 2:

Thus, if a = 9 and d = – 5

Then the three terms are,

First term a – d = 9 – (– 5) = 9 + 5 = 14

Second term a = 9

Third term a + d = 9 + (– 5) = 9 – 5 = 4

Thus, the A.P. is 14, 9, 4