已知两个等差数列{an}{bn},其前n项的和分别为Sn和Tn,并且Sn/Tn=7n+2/n+3,则a7/b8=
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已知两个等差数列{an}{bn},其前n项的和分别为Sn和Tn,并且Sn/Tn=7n+2/n+3,则a7/b8=
sn=(n/2)(a1+an),Tn=(n/2)(b1+bn),设an公差为d1,bn公差为d2
Sn/Tn=(a1+an)/(b1+bn)=(nd1+a1-d1)/(nd2+b1-d2)=(7n+2)/(n+3)
令d2=m,m≠0,则d1=7m,a1-d1=2m,b1-d2=3m
得a1=9m,b1=4m
a7=a1+6d1=9m+42m=51m
b8=b1+7d2=4m+7m=11m
a7/b8=51m/11m=51/11 答案是31/6
Sn/Tn=(a1+an)/(b1+bn)=(nd1+a1-d1)/(nd2+b1-d2)=(7n+2)/(n+3)
令d2=m,m≠0,则d1=7m,a1-d1=2m,b1-d2=3m
得a1=9m,b1=4m
a7=a1+6d1=9m+42m=51m
b8=b1+7d2=4m+7m=11m
a7/b8=51m/11m=51/11 答案是31/6
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