Question

find the four numbers in A.P such that the sum is 38 and the product is 2856


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Answer 1

It has given that, four consecutive numbers are given in AP where the sum of numbers is 38 and the product of numbers is 2856.

solution : let (a - 3d), (a - d) , (a + d) and (a + 3d) are four numbers in ap.

sum = 38

⇒(a - 3d) + (a - d) + (a + d) + (a + 3d) = 38

⇒4a = 38

⇒a = 9.5 ......(1)

Product= 2856

⇒(a - 3d)(a - d)(a + d)(a + 3d) = 2856

⇒(a² - 9d²)(a² - d²) = 2856

⇒(9.5² - 9d²)(9.5² - d²) = 2856

⇒(9.5)⁴ - (9.5)²10d² + 9d⁴ = 2856

⇒8145.0625 - 902.5d² + 9d⁴ = 2856

⇒9d⁴ - 902.5d² - 5289.0625 = 0

⇒d =-9.7 , 9.7 , 2.5 or -2.5

now you can find out the numbers using value of a and d.

Let d = 2.5 and a = 9.5

So, (a - 3d) = 9.5 - 7.5 = 2

(a - d) = 9.5 - 2.5 = 7

(a + d) = 9.5 +2.5 =12

(a + 3d) = 9.5 +7.5 = 17

Numbers are 2, 7, 12 and 17