设数列{an}，{bn}都是正项等比数列，Sn，Tn分别为数列{lgan}与{lgbn}的前n项和，且SnTn＝n2n+1，则logb5a5=919919．

Question

设数列{a_n}，{b_n}都是正项等比数列，S_n，T_n分别为数列{lga_n}与{lgb_n}的前n项和，且(Sn)/(Tn)＝(n)/(2n+1)，则logb5a5=

(9)/(19)

(9)/(19)

．

Answer 1

设正项等比数列{a_n}的公比为q，设正项等比数列{b_n}的公比为p，则数列{lga_n}是等差数列，公差为lgq，{lgb_n}是等差数列，公差为lgp．
故 S_n =n•lga₁+(n(n−1))/(2)• lgq，同理可得 T_n =n•lgb₁+(n(n−1))/(2)• lgp．
又(Sn)/(Tn)＝(n)/(2n+1)=(lga1+(n−1)/(2)lgq)/(lgb1+(n−1)/(2)lgp)，
∴logb5a5=(lga5)/(lgb5)=(lga1+4lgq)/(lgb1+4lgp)=(S9)/(T9)=(9)/(19)，
故答案为 (9)/(19)．