Question

已知公比为q的等比数列{an}前六项和为S6=21,且4a1,3/2a2,a2,成等差数列,(1)求an

Answer 1

(1)因为4a1,3/2a2,a2,成等差数列

所以2*（3/2a2）=4a1+a2------>q=2

S6=[a1(1-q^6)]/(1-q)=21----->a1=1/3

an=a1q^(n-1)=(1/3)*2^(n-1)

(2)由（1）得a1=1/3

所以｛bn}的公差为-1/3

bn=2-(n-1)(-1/3)

Tn=2n-n(n-1)/6＞2----->n^2-13n+121