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If $a,b,c$ are in AP. $b-a,c-b$ and a are in GP. then $a:b:c$ is
\[\begin {array} {1 1} (a)\;1:2:3 & \quad (b)\;1:3:5 \\ (c)\;2:3:4 & \quad (d)\;1:2:4 \end {array}\]
jeemain
eamcet
math
2007
q28
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Oct 24, 2013
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meena.p
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(a) 1:2:3
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Nov 7, 2013
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pady_1
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