Question

一道数学数列题

在等差数列{an}中,若a2+a5+a8=9,a3*a5*a7=-21,求数列的通项公式

Answer 1

a2+a5+a8=9,
3a5=9
a5=3
a3*a5*a7=-21
(a5-2d)a5(a5+2d)=-21
[(a5)^2-4d^2]a5=-21
[3^2-4d^2]*3=-21
3^2-4d^2=-7
4d^2=16
d^2=4
d=±2
当d=2时
a5=a1+4d
3=a1+4*2
a1=-5
an=a1+(n-1)d
=-5+2(n-1)
=2n-7
当d=-2时
a5=a1+4d
3=a1+4*(-2)
a1=11
an=a1+(n-1)d
=11-2(n-1)
=13-2n O(∩_∩)O