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设数列{an}是等差数列,a1=1,Sn=a1+a2+…+an,数列{bn}是等比数列,Tn=b1+b2+…+bn,若a3=b2,S5=2T2-6,且Tn=9.
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅱ)当自然数n取何值时,Sn>Tn?
设等差数列{an}的前n项和是Sn,且a1=10,a2=9,那么下列不等式中成立的是
A.a10-a11<0
B.a20+a22>0
C.S20-S21<0
D.S40+a41<0