Question

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

Answer 1

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

by the given conditions

(a – 3d)(a + 3d) : (a – d)(a + d) = 2 : 3

⇒ (a^2 – 9d^2) : (a^2 – d^2) = 2 : 3

⇒ 3(a^2 – 9d^2) = 2(a^2 – d^2)

⇒ 3a^2 – 27d^2 = 2a^2 – 2d^2

⇒ 3a^2 – 2a^2 = 27d^2 – 2d^2

⇒ a^2 = 25d^2

⇒ 52 = 25d^2

⇒ 25 = 25d^2

⇒ d^2 = 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.

I hope this answers will help you