設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1
設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1
數(shù)學(xué)人氣:377 ℃時間:2020-05-25 17:06:41
優(yōu)質(zhì)解答
S4=4a1+(4x3)/2d=-62S6=6a1+(6x5)/2d=-75d=3,a1=-20∴an=a1+(n-1)d=3n-23 由an>0即3n-23>0得 n>23/3所以當(dāng)n0;設(shè)Tn為{|an|}的前n項(xiàng)和當(dāng)n=8時,Tn=|a1|+|a2|+|a3|+…+|a7|+|a8|+…+|an|=-a1-a2-a3-…-a7+a8+…+an=Sn...
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