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Question

Three distinct real numbers a, b, c are in G .P . such that a+b+c=xb, 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 A x < -1 or x > 3
a+ar+ar2=x.ar
or r2+r(1x)+1=0 (r is real)
Δ>0 i.e. (1x)24>0
or, x22x3>0
or, (x+1)(x3)>0
x<1 or x>3

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