Question

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

Answer 1

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