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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 21:38:24 IST
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Find the largest term common in sequences 1,11,21,31,41,..... to 100 terms and 31,36,41,46,51,56,.... to 100 terms ????????
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Umang |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 22:04:38 IST
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is the answer 521???
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 22:07:03 IST
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Hey ! actually , it aws an objective question . and the answer given is 471 . Pls explain how u got ur answer !!!!!
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Umang |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 22:14:35 IST
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in the first series 1,11.... the 100th term is 991 (by the formula Tn=a+(n-1)d) in the second series31,36,41............100th term is 526. first series is 1,11..........521.......991 second series is 31,36..........516,521,526 hence the largest common term is 521. is there any mistake in my solution?? please let me know..
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 22:30:33 IST
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reply soon!!!!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 22:52:22 IST
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common differences of first series= 10 common differences of second series= 5 first common term = 41 if you observe closely there will be a sequence for common terms with common difference 10*5=50 hence the sequence will be 41, 91, 141, 191 ..................... as saarika found it last terms of the two sequences are 991,and 526. take the minimum one i.e 526 41+(n-1)50=526 n=10.7 n is a integer so n=10 putting in the formula t 10 = 491 i think 471 is not the answer as it does not satisfy the equation if 491 in the option!!!! please verify
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Shashank nayak |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 23:10:25 IST
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03/03/2007 22:52 Subject: Re:common term | | | common differences of first series= 10 common differences of second series= 5 first common term = 31 if you observe closely there will be a sequence for common terms with common difference 10 hence the sequence will be 31,41,51..................... as saarika found it last terms of the two sequences are 991,and 526. take the minimum one i.e 526 31+(n-1)10=526 n=50.5 n is a integer so n=50 putting in the formula t 10 = 521 sorrrrrrrrrrrrry!!!!!!!!! 521 is right saarika is right ignore the first post |
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Shashank nayak |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Mar 2007 11:29:22 IST
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521 will be the answer, see it can be solved logically also, the first series's numbers ranging frm [1,526) all will be also of that of the other series, so defintely 521 will be the answer, hey umang,hav u taken this question frm fiitjee package??? its given thr also as 471... but the answer is defintely 521!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Mar 2007 12:38:05 IST
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hey ,the answer is 521.bcoz 521 occurs in both the series which is of course >471 i do not care what the book says bcoz if u can see it with ur eyes that it is 521,why should we say that we are wrong and the book is right ????
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Mar 2007 21:10:25 IST
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Hey shashank !! yes , ur solution is correct . And sarika ,you hav applied hit and trial method , though ur answer is correct. I dont know why the book says the answer=471 , but yes , 521 must be the correct answer !!!!!!!
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Umang |
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