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Question

If x,y,z are three consecutive terms of a GP, where x>0 and the common ratio is r, then the inequality z+3x>3y holds for

A
r(,1)
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B
r=245
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C
r(3,+)
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D
r=12
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Solution

The correct options are
A r(,1)
B r=245
C r(3,+)
D r=12
x>0
y=rx,z=r2x
z+3x>3y
=>r2x+3x>3rx
=>x(r23r+3)>0
(r23r+3)>0
(r32)2+34>0
which is true for all R.
Hence, options A,B,C,D are correct.

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