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

Also, Given product of three consecutive term is 504

⇒ (a – d)× a × (a + d) = 504

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

⇒ 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