Question

数学数列题

设等差数列{an}前n项和为Sn，若a6=S3=12，则{an}的通项an=

求过程，谢谢了

Answer 1

a6 = a1 + 5d = 12 (1)
S3 = a1 + a1 + d + a1 + 2d = 3a1 + 3d = 12
a1 + d = 4 (1)
解(1)(2), a1 = 2, d = 2
an = 2 + 2(n-1)

Answer 2

a6=S3=12
a6=a1+5d=12 (1)
s3=3a1+3d=12 (2)
(1)×3-(2)得 12d=24
d=2
把d=2代入(1)得 a1=2
an=a1+(n-1)d=2+2(n-1)=2n