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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If the numbers $\;1,x^2,6-x^2\;$ are in GP , then value of $x$ is :

$(a)\;\pm\sqrt{3}\qquad(b)\;\pm3\qquad(c)\;\pm\sqrt{2}\qquad(d)\;\pm2$

Can you answer this question?
 
 

1 Answer

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Answer : (c) $\;\pm\sqrt{2}$
Explanation : For GP
$(x^2)^2=(1)\;(6-x^2)$
$x^4+x^2-6=0$
$(x^2+3)(x^2-2)=0$
$x^2=2$
$x=\pm\sqrt{2}.$
answered Jan 21, 2014 by yamini.v
 

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