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If-S-n-sum-of-sequence-2n-2-n-Find-the-value-of-the-1002-term-




Question Number 7656 by Tawakalitu. last updated on 07/Sep/16
If  S_n  (sum of sequence) = 2n^2  + n  Find the value of the 1002 term
IfSn(sumofsequence)=2n2+nFindthevalueofthe1002term
Commented by prakash jain last updated on 12/Sep/16
n^(th)  term T_n =S_n −S_(n−1)   =2n^2 +n−2(n−1)^2 −(n−1)  =2n^2 +n−2n^2 +4n−2−n+1  =4n−1  T_(1002) =4×1002−1=4007
nthtermTn=SnSn1=2n2+n2(n1)2(n1)=2n2+n2n2+4n2n+1=4n1T1002=4×10021=4007
Answered by Rasheed Soomro last updated on 09/Sep/16
S_n =2n^2 +n  t_(1002) =?  Let a and d are first term  and  common difference  of the AP.  S_n =(n/2)[2a+(n−1)d]  Let first determine sum of 2003 terms,which  will help us to determine  t_(1002)  .  S_(2003) =((2003)/2)[2a+(2003−1)d]=2(2003)^2 +2003            =[2a+2002d]=((2(2003)^2 +2003)/(2003))×2           2[a+1001d]=[2(2003)+1]×2           a+(1002−1)d=(([2(2003)+1]×2)/2)=4006+1=4007          t_(1002) =4007
Sn=2n2+nt1002=?LetaanddarefirsttermandcommondifferenceoftheAP.Sn=n2[2a+(n1)d]Letfirstdeterminesumof2003terms,whichwillhelpustodeterminet1002.S2003=20032[2a+(20031)d]=2(2003)2+2003=[2a+2002d]=2(2003)2+20032003×22[a+1001d]=[2(2003)+1]×2a+(10021)d=[2(2003)+1]×22=4006+1=4007t1002=4007
Commented by Tawakalitu. last updated on 09/Sep/16
Thanks but they wrote 4007 here as answer. maybe they are wrong  thanks sir.
Thanksbuttheywrote4007hereasanswer.maybetheyarewrongthankssir.
Commented by Rasheed Soomro last updated on 09/Sep/16
They are right. There was a mistake in my answer.  I had multiplied by 2 instead of divided in last two steps.  I have corrected it now. Sorry for mistake!
Theyareright.Therewasamistakeinmyanswer.Ihadmultipliedby2insteadofdividedinlasttwosteps.Ihavecorrecteditnow.Sorryformistake!
Commented by Tawakalitu. last updated on 11/Sep/16
Thanks so much. i really appreciate.
Thankssomuch.ireallyappreciate.
Commented by Tawakalitu. last updated on 11/Sep/16
Thanks so much. i really appreciate.
Thankssomuch.ireallyappreciate.

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