In an A.P. sum of three consecutive terms is 27 and their product is 504 find the terms? (Assume that three consecutive terms in A.P. are a – d, a, a + d.)

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
⇒ 9² – d² = 56 (since, (a – b) (a + b) = a² – b²)
⇒ 81 – d² = 56
⇒ d² = 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