两个等差数列{an},{bn}的前n项之和分别为An和Bn,且An/Bn = (7n+2)/(7n+3),求a7/b7的值。
来源:百度知道 编辑:UC知道 时间:2024/05/16 21:38:57
两个等差数列{an},{bn}的前n项之和分别为An和Bn,且An/Bn = (7n+2)/(7n+3),求a7/b7的值。
加油 加油~!>.<
加油 加油~!>.<
An/Bn =[na1+n(n-1)d/2]/(nb1+n(n-1)d'/2]
=[2a1+(n-1)d]/[2b1+(n-1)d']
=[dn+(2a1-d)]/[d'n+(2b1-d)]
=(7n+2)/(7n+3)
所以d=d'=7,a1=9/2,b1=5
a7/b7=(a1+6d)/(b1+6d')=46/47
An/Bn = (7n+2)/(7n+3)
A6/B6=(42+2)/(42+3)=44/45
A7/B7=(49+2)/(49+3)=51/52
a7=A7-A6
b7=B7-B6
a7/b7=(A7-A6)/(B7-B6)=
已知等差数列{an},{bn}...
数列{an}为等比数列,{bn}为等差数列,
若{an}和{bn}数列是等差数列,求证{an+bn}也是等差数列.
已知{an}是等差数列,bn=kan+m(k,m为常数).求证{bn}是等差数列
已知两个等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn:Tn=2n:(3n+1),则用n表示an/bn=,,怎么做的?
等差数列{an}的前n项和Sn=an^2+bn+c
已知数列{An}满足An=n(n+1)^2,请问是否存在等差数列{Bn},使
已知等差数列{an},an=21-2n,由知bn=|an|,求数列{bn}的前30项和
等差数列{an} {bn}的前n项的分别为Sn Tn。
等差数列{an},{bn},的前n项和分别为sn,tn,