设等差数列{an}是递增数列,前n项和为sn,且a1,a3,a9.成等比数列,S5=a5*2.求数列{an}的通项公式快

Question

设等差数列{an}是递增数列,前n项和为sn,且a1,a3,a9.成等比数列,S5=a5*2.求数列{an}的通项公式
快

Answer 1

an = a1 + d
a1*m = (a1+2d)
a1(m-1) = 2d
(a1+2d)*m = (a1 + 8d)
a1(m-1) = 8d - 2dm
8d -2dm = 2d
2dm = 6d
m = 3
3a1 = a1+2d
d = a1
s1 = a1
s3 = 3a1
s9 = 9a1
s5 = a1+2a1+3a1+4a1+5a1 = 15a1 = (5a1)^2
a1 = 3/5
an = 3n/5