设等差数列｛an｝的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1_数学解答

设等差数列｛an｝的前n项和为Sn,S4=-62,S6=-75,求︳a1︳+︳a2︳+︳a3︳+…+︳a1

人气：310 ℃　时间：2020-05-25 17:06:41

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所以当n0；设Tn为{|an|}的前n项和当n=8时,Tn=|a1|+|a2|+|a3|+…+|a7|+|a8|+…+|an|=-a1-a2-a3-…-a7+a8+…+an=Sn...