Question

divide 20 into four parts which are in AP such that the product of the first and fourth is to the product of the second and third is 2:3

Answer 1

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let the four numbers to be the a-3d ,a-d, a+d ,a+3d as the numbers are in arithmetic progressions with common difference: 2d and (a-3d) +(a-d)+(a+d) +(a+3d)=20 we get 4a=20 a=5 (a-3d)(a+3d)/(a+d)(a-d)=2/3 (a-3d)(a+3d)*3=(a+d)(a-d)*2 i.e. 3a^2-27d^2=2a^2-2d^2 a^2=25d^2 :now substitute a=5 we get 25=25d^2 d^2=1 d=+1,-1 so the numbers become 2,4,6,8

Answer 2

Answer:

Let the numbers be 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