S19/T19=[(a1+a19)*19/2]/[(b1+b19)*19/2]
=(a1+a19)/(b1+b19)
=2a10/2b10=a10/b10
=(5*19+2)/(7*19-1)
=97/132
a10/b10=97/132
若等差数列{an}与{bn}的前n项之和分别为Sn和Tn,若Sn/Tn=(5n+2)/(7n-1),则a10/b10=
1个回答
相关问题
-
等差数列｛an｝,｛bn｝的前n项分别为Sn,Tn,若Sn/Tn=2n/3n+1,则an/bn=多少?
-
已知{an},{bn}均为等差数列,其前n项之和分别为Sn、 Tn,若Sn/Tn=（2n+2）/（n+3）,求a10/b
-
等差数列an`bn`的前n项和分别为Sn`Tn.若Sn/Tn=(7n+1)/(4n+27),求an/bn
-
等差数列{an}、{bn}的前n项和分别为Sn、Tn,若Sn/Tn=2n/3n+1,求an/bn
-
等差数列{An},{Bn}的前n项和为Sn与Tn,若Sn/Tn=2n/3n+1,则A5/B7的值是
-
等差数列{an}{bn}的前n项和分别为Sn,Tn,若Sn/Tn=2n/3n+1,求a5/b5=多少
-
...求正解等差数列{an}、{bn}的前n项和分别为Sn、Tn,若Sn/Tn=2n+2/n+3,则 a7/b7=?(_
-
设Sn ,Tn分别为等差数列{an}{bn}的前n项和,Sn/Tn=(5n-2)/(n+3),则an/bn=?
-
等差数列{an},{bn}的前n项和分别为Sn,Tn,Sn/Tn=2n/3n+1则a3/b5
-
等差数列{an} {bn}的前n项的分别为Sn Tn.若Sn/Tn=2n/（3n+1）,则求an/bn的极限