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Find three consecutive numbers in G.P. whose sum is 28 and product is 512.

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Question

Find three consecutive numbers in G.P. whose sum is 28 and product is 512.

Sum
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Solution

`a/r`, a, ar

`a/r` +  a + ar = 28

`a(1/r + 1 + r)` = 28

`a/r xx a xx ar` = 512

⇒ a3 = 512

a = 8

`8(1/r + 1 + r)` = 28

2(1 + r + r2) = 7r

2r2 + 2r − 7r + 2 = 0

2r2 − 5r + 2 = 0

2r2 − 4r − r + 2 = 0

2r(r − 2) − 1(r − 2) = 0

(2r − 1)(r − 2) = 0

r = `1/2`, 2

1. r = `1/2`, a = 8

= `8/(1/2), 8, 8(1/2)`

= 16, 8, 4

2. r = 2, a = 8

`8/2, 8, 8(2)`

= 4, 8, 16

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Chapter 9: Arithmetic and geometric progression - Exercise 9D [Page 194]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 9 Arithmetic and geometric progression
Exercise 9D | Q 23. | Page 194
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