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Question

If a,b and c be three distinct real numbers in G.P. and a+b+c=xb, then x cannot be

A
4
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B
3
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C
2
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D
2
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Solution

The correct option is D 2
br,b,brG.P.(|r|1)

given a+b+c=xb

br+b+br=xb

b=0 (not possible)

or 1+r+1r=xx1=r+1r

x1>2 or x1<2

x>3 or x<1

So x can't be 2.

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