等差数列{an}中,3a1+2a5=21.2a4=a3+a6-2其前n项和为sn求数列{an}的通项公式

2024-12-20 16:15:51
推荐回答(3个)
回答1:

设公差为d
因为3a1+2a5=21,2a4=a3+a6-2
所以3a1+2(a1+4d)=21
2(a1+3d)=a1+2d+a1+5d-2
a1= 1 d=2
所以an=a1+(n-1)d=2n-1

回答2:

3a1+2a5=21.2a4=a3+a6-2
5a1+8d=21,2a1+6d=2a1+7d-2
5a1+8d=21,d=2
a1=1,d=2
an=1+2(n-1)=2n-1
an=2n-1

回答3:

an = a1+(n-1)d
3a1+2a5 = 5a1+8d =21 (1)
2a4=a3+a6-2
2a1+6d = 2a1+7d-2
d=2 (2)
sub (2) into (1)
5a1+16=21
a1=5
an = 5+(n-1)2 = 2n+3