Question

divide 20 into 4 parts which are in AP such that the product of the first and the fourth terms is to the product of the second and third term in the ratio 2:8​

Answer 1

Answer:

Step-by-step explanation:

Let the four parts be a – 3d, a – d, a + d and a +3d

Hence (a – 3d) + (a – d) + (a + d) + (a +3d) = 20

⇒ 4a = 20

∴ a = 5

It is also given that (a – 3d)(a + 3d) : (a – d)(a + d) = 2 : 3

⇒ (a2 – 9d2) : (a2 – d2) = 2 : 3

 

⇒ 3(a2 – 9d2) = 2(a2 – d2)

⇒ 3a2 – 27d2 = 2a2 – 2d2

⇒ 3a2 – 2a2 = 27d2 – 2d2

⇒ a2 = 25d2

⇒ 52 = 25d2

⇒ 25 = 25d2

⇒ d2 = 1

∴ d = ± 1

Case (i): If d = 1

Hence (a – 3d) = (5 – 3) = 2

(a – d) = (5 – 1) = 4

(a + d) = (5 + 1) = 6

(a + 3d) = (5 + 3) = 8

Hence the four numbers are 2, 4, 6 and 8.

Case (ii): If d = –1

Hence (a – 3d) = (5 + 3) = 8

(a – d) = (5 + 1) = 6

(a + d) = (5 – 1) = 4

(a + 3d) = (5 – 3) = 2

Hence the four numbers are 8, 6, 4 and 2.

hope it works