Question

等差数列公差为d，a1>=1，a24>=24，S12<=168，则a9-d^2的取值范围是多少

Answer 1

等差数列{an}中，a1>=1,S12=6(a1+a12)<=168,
∴a12<=28-a1,a12<=27,
12d=a24-a12>=24-27=-3,d>=-1/4;
11d=a12-a1<=26,d<=26/11:
∴-1/4<=d<=26/11,
a9-d^2=a1+8d-d^2=-(d-4)^2+16+a1是d的增函数，
d=-1/4时a9-d^2>=-17/16,
a9-d^2=a12-3d-d^2=a12-(d+3/2)^2+9/4是d的减函数，
d=-1/4时,a9-d^2<=27+3/4-1/16=443/16.
∴a9-d^2的取值范围是[-17/16,443/16].

Answer 2

为什么我们老师说是[8,249/16].，我也不晓得怎么算，有人有过程啊。