1.
(1)∵數(shù)列{an}是等差數(shù)列
a1+d=1,a1+4d=-5
解得 d=-2,a1=3
∴an=3-2(n-1)=-2n+5
(2)
an=5-2n≥0 ==> n≤5/2 ==>n≤2
∴a1>0,a2>0,a3 n=50
4.∵等差數(shù)列{an}中,
a1+a2+a3+a4+a5=20
∴a1+a5=a2+a4=2a3
∴5a3=20,a3=4
5.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,若a4=18-a5則S8=多少?
∵a4=18-a5 ∴a4+a5=18 ∴a1+a8=a4+a5=18
∴S8=(a1+a8)*8/2=8*8/2=32
6.1∵{an} 公差為-2,
a1+a4+a7+……+a16=498
∴ a3+a6+a9+……+a18
=( a1+2a)+(a4+2d)+(a7+2d)+.+(a16+2d)
=(a1+a4+a7+.+a16)+6×2d
=498-24
=464