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{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Sn/Tn=n/(2n+1),则logb5a5=?(b5a5是真数)

题目详情
{an },{bn }都是正项等比数列,Sn,Tn分别为数列{lg an }与{lg bn }的前n项和,且Sn/Tn=n/(2n+1),则log b5a5=?(b5a5是真数)
▼优质解答
答案和解析
{an}{bn}成等比数列,则{lgan},{lgbn}成等差数列,
Sn=lga1+lga2+lga3+...+lgan=nlga1+n(n-1)d1/2
Tn=lgb1+lgb2+lgb3+...+lgbn=nlgb1+n(n-1)d2/2
Sn/Tn=[nlga1+n(n-1)d1/2]/[nlgb1+n(n-1)d2/2]
=[lga1+(n-1)d1/2]/[lgb1+(n-1)d2/2]=n/(2n+1)
令:n=9,上式为;
[lga1+4d1]/[lgb1+(4d2/]=9/19,即:
lga5/lgb5=9/19,由对数换底公式得:
lga5/lgb5=logb5(a5)=9/19