Question

18. Divide 20 into four parts which are in A. P. such that the product of
the first and fourth is to the product of the second and third is 2:3.
Type IV.​

Answer 1

Answer:

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

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