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Q-1-which-term-of-the-sequence-2005-2000-1995-1990-1985-is-the-first-negative-term-plese-give-answer-Q-2-for-an-A-P-show-that-t-m-t-2n-m-2t-m-n-give-answer-Q-3-fin

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Question Number 7654 by Rohit last updated on 07/Sep/16
Q.1 which term of the sequence 2005, 2000,1995,1990,1985,........................ is the first negative term. plese give answer Q.2 for an A.P. show that t_m +t_(2n+m) = 2t_(m+n) give answer Q.3 find the maximum sum of the A.P. 40+38+36+34+32+........ give answer plese sir
Q.1whichtermofthesequence2005,2000,1995,1990,1985,……………………isthefirstnegativeterm.plesegiveanswerQ.2foranA.P.showthattm+t2n+m=2tm+ngiveanswerQ.3findthemaximumsumoftheA.P.40+38+36+34+32+……..giveanswerplesesir
Answered by Rasheed Soomro last updated on 07/Sep/16
Q.2 for an A.P. show that t_m +t_(2n+m) =2t_(m+n) −−−−−−−−−−−−−−−−−−−− Let the first term and common differnce are a and d respectively. t_m =a+(m−1)d t_(2n+m) =a+(2n+m−1)d t_(m+n) =a+(m+n−1)d LHS: t_m +t_(2n+m) =( a+(m−1)d )+( a+(2n+m−1)d ) =2a+(m−1+2n+m−1)d =2a+(2m+2n−2)d =2[ a+(m+n−1)d ] =2 t_(m+n) =RHS Proved.
Q.2foranA.P.showthattm+t2n+m=2tm+n−−−−−−−−−−−−−−−−−−−−Letthefirsttermandcommondiffernceareaanddrespectively.tm=a+(m−1)dt2n+m=a+(2n+m−1)dtm+n=a+(m+n−1)dLHS:tm+t2n+m=(a+(m−1)d)+(a+(2n+m−1)d)=2a+(m−1+2n+m−1)d=2a+(2m+2n−2)d=2[a+(m+n−1)d]=2tm+n=RHSProved.
Answered by Rasheed Soomro last updated on 08/Sep/16
Q.3 Maximum sum of 40+38+36+34+32+........ −−−−−−−−−−−−−−−− Maximum sum is 40+38+36+......+2 Because the next term 0 doesn′t affect the sum and after it are negative terms which decrease the sum. l=last term=40+(n−1)(−2)=2 [n is number of terms ] 40−2n+2=2 n=20 S_n =(n/2)(a+l) S_(20) =((20)/2)(40+2)=420 MAXimum Sum is 420
Q.3Maximumsumof40+38+36+34+32+……..−−−−−−−−−−−−−−−−Maximumsumis40+38+36+……+2Becausethenextterm0doesn′taffectthesumandafteritarenegativetermswhichdecreasethesum.l=lastterm=40+(n−1)(−2)=2[nisnumberofterms]40−2n+2=2n=20Sn=n2(a+l)S20=202(40+2)=420MAXimumSumis420
Commented by Rohit last updated on 08/Sep/16
kese aya 40+38+36.....2
keseaya40+38+36…..2
Commented by Rasheed Soomro last updated on 08/Sep/16
The series contains even numbers: 40,38,36....So last positive term is even number 2 After that is 0,which doesn′t affect the sum and after 0 negative even numbers come,which will cause to decrease the sum.S maximum sum is 40+38+36+...+2. Upto 2 the sum is increasing.
Theseriescontainsevennumbers:40,38,36….Solastpositivetermisevennumber2Afterthatis0,whichdoesn′taffectthesumandafter0negativeevennumberscome,whichwillcausetodecreasethesum.Smaximumsumis40+38+36+…+2.Upto2thesumisincreasing.
Answered by Rasheed Soomro last updated on 08/Sep/16
Q.1 which term of the sequence 2005, 2000,1995,1990,1985,....................... is the first negative term. −−−−−−−−−−−−−−−−−−−−−−− Last non-negative term=0 Let 0 is nth term,a first term,d common difference. nth term=a+(n−1)d =2005+(n−1)(2000−2005)=0 2005−5n+5=0 n=2010/5=402 Last non-negative term=402nd term 403rd term is first negative term.
Q.1whichtermofthesequence2005,2000,1995,1990,1985,…………………..isthefirstnegativeterm.−−−−−−−−−−−−−−−−−−−−−−−Lastnon−negativeterm=0Let0isnthterm,afirstterm,dcommondifference.nthterm=a+(n−1)d=2005+(n−1)(2000−2005)=02005−5n+5=0n=2010/5=402Lastnon−negativeterm=402ndterm403rdtermisfirstnegativeterm.

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