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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 May 2007 17:21:54 IST
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i faintly remember this question which came in I.P. university exam do correct me if wrong Q. if a , b , c are in AP then (1+a ) , ( 2+ b ) , (1 + c) are in :: (a) AP (b) GP (C) HP (D) NOT i don't remember exactly.......somebody knows the full question please
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 May 2007 17:34:50 IST
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Hi. Here is a shortcut. Take a,b,c as 1,2,3 1+a=2 2+b=4 1+c=4 since 2,4,4 are neither in AP,GP or HP the ans is (D)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 May 2007 23:04:15 IST
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Excellent solution joy.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 May 2007 09:33:06 IST
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thanx yaar joyfrancis
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 May 2007 10:18:38 IST
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sorry joy but a,b,c r in AP. its given but 1,2,4 r in no case in AP.
the solution is still d). u can check by the method provided by joy
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Being beaten is often a temporary condition, giving up is what makes it permanent. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 May 2007 10:27:39 IST
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hey Joy howz 1,2,4 in ap?? See, since a,b,c are in AP write b=(a+c)/2 the sereis becomes 1+a,2+(a+c)/2,1+c now this could be veriofied as none of ap,gp,hp..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 May 2007 12:42:13 IST
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oh yeah, i took it as a GP. See the edited version
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