數列{an},{bn}都是正項等比數列,通項分別為q^(n-1),p^(n-1),則數列{lgan}與{lgbn}通項為(n-1)lgq和(n-1)lgp是等差數列,Sn=(n-1)²/2*lgq,Tn=(n-1)²/2*lgp,Sn/Tn=lgq/lgp=n/2n+1,則lga5/lgb5=5/11sorry啊，答案是9/19，麻煩您能再算一遍，可以嗎？數列{an},{bn}都是正項等比數列,通項分別為a1*q^(n-1),b1*p^(n-1),則數列{lgan}與{lgbn}通項為lga1+(n-1)lgq和lgb1+(n-1)lgp是等差數列，公差分別為lgq和lgp，Sn=nlga1+n(n-1)lgq/2，Tn=nlgb1+n(n-1)lgp/2，Sn/Tn=[lga1+(n-1)lgq/2]/[lgb1+(n-1)lgp/2]=n/2n+1，lga5/lgb5=(lga1+4lgq)/(lgb1+4lgp),Sn/Tn=[lga1+(n-1)lgq/2]/[lgb1+(n-1)lgp/2]=n/2n+1當(n-1)/2=4時，即n=9，則Sn/Tn=(lga1+4lgq)/(lgb1+4lgp)=n/2n+1=9/19