Question

In an A.P sum of three consecutive terms is 27 and their products is 504 find the terms Assume that three consecutive terms in A.P are a-d, a, a+d

Answer 1

Answer:

-4 ,9,24

Step-by-step explanation:

a-d+a+a+d=27

3a=27

a=27/3 =9

(a-d)(a)(a+d)=504

(a^3)-(d^2)=504

(9^3)-(d^2)=504

729-504 = (d^2)

225=(d^2)

d=✓225

d=15

a-d = 9-15 = -4

a = 9

a+d = 9+15 = 24

Answer 2

Let the consecutive terms are a - d, a, a + d.

Their sum = 27

⇒ (a - d) + a + (a + d) = 27

⇒ 3a = 27      ⇒ a = 27/3 = 9

Their product = 504

⇒ (a - d)(a)(a + d) = 504

⇒ (9 - d)(9)(9 + d) = 504    {a=9}

⇒ (9 - d)(9 + d) = 504/9

⇒ 9² - d² = 56    ⇒ 81 - 56 = d²

⇒ 25 = d²            ⇒ ± 5 = d   

∴ terms are 9 - 5, 9, 9 + 5 or 9 - (-5), 9, 9 + (-5)

⇒ 4, 9, 14   or 14, 9 , 4