Question

已知等差数列(an)的公差为2，若a1,a3,a4成等比则a2=?

Answer 1

a3=a1+2d=a1+4
a4=a1+3d=a1+6
因为a1,a3,a4成等比数列,则
a4/a3=a3/a1
(a1+4)^2=a1(a1+6)
解之,a1=-8
则 a2=a1+d=-8+2
=-6

Answer 2

a1=a2-2,a3=a2+2,a4=a2+4
a1,a3,a4成等比 => a3/a1=a4/a3
即(a2+2)/(a2-2)=(a2+4)/(a2+2)
解得a2=-6

Answer 3

a1=a2-2,a3=a2+2,a4=a2+4
a1,a3,a4成等比 => a3/a1=a4/a3
即(a2+2)/(a2-2)=(a2+4)/(a2+2)
解得a2=-6

Answer 4

选C啊 

Answer 5

(1)a3=a1+2d=a1+4
a4=a1+3d=a1+6
a3
^2=a1a4
(a1+4)^2=a1(a1+6)
a1=-8
a2=a1+2=-6
(2)在等比数列[an}中,s2,(s4-s2),(s6-s4)仍成等比数列，
∴
(s4-7)²=7(91-s4)，即(s4)²-7(s4)-588=0,
∴
s4=28,(舍去s4=-21,
∵
此时由s2/s4=1/(q²+1)得q²<0).