问题描述:
已知{an}是等差数列,Sn是其前n项和,(1)a2=-1,S15=75,求an与Sn;
(2)a1+a2+a3+a4=124,an+an-1+an-2+an-3=156,Sn=210,求项数n.
最佳答案:
(1)∵{an}是等差数列,Sn是其前n项和,a2=-1,S15=75,∴a2=a1+d=-1S15=15a1+15×142d=75,解得a1=-2,d=1,∴an=-2+(n-1)×1=n-3.Sn=-2n+n(n-1)2×1=n2-5n2.(2)∵{an}是等差数列,Sn是其前n项和,a1+a2+a3+...