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Question

Three distinct real numbers, a, b, c are in G.P. such that a + b+ c= x b, then


A

0<x<1

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B

-1<x<3

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C

x<-1 or x>3

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D

-1<x<2

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Solution

The correct option is C

x<-1 or x>3


a + ar + ar2 = x. ar

or, r2 + r (1-x) + 1 = 0, r is real

∆>0 i.e.(1x)2- 4 >0

Or,x2-2x-3>0

Or, (x+1) (x-3)>0

x<-1 or x>3


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