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Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7. - Mathematics

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Question

Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.

Sum

Solution

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

Then, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 96

⇒ 4a = 96

⇒ a = 24

It is given that 

`\implies ((a - d)(a + d))/((a - 3d)(a + 3d)) = 15/7`

`\implies (a^2 - d^2)/(a^2 - 9d^2) = 15/7`

`implies ( 576 - d^2)/(576 - 9d^2) = 15/7`

`\implies` 4032 – 7d2 = 8640 – 135d2

`\implies` 128d2 = 4608

`\implies` d2 = 36

`\implies` d = ±6

When a = 24, d = 6

a – 3d = 24 – 3(6) = 6 

a – d = 24 – 6 = 18

a + d = 24 + 6 = 30

a + 3d = 24 + 3(6) = 42

When a = 24, d = –6

a – 3d = 24 – 3(–6) = 42

a – d = 24 – (–6) = 30

a + d = 24 + (–6) = 18

a + 3d = 24 + 3(–6) = 6

Thus, the four parts are (6, 18, 30, 42) or (42, 30, 18, 6).

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Simple Applications of Arithmetic Progression
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Chapter 10: Arithmetic Progression - Exercise 10 (D) [Page 146]

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Selina Mathematics [English] Class 10 ICSE
Chapter 10 Arithmetic Progression
Exercise 10 (D) | Q 4 | Page 146
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