已知两个等差数列{an},{bn},其前n项和分别为Sn,Tn,且Sn/Tn=7n+2/n+3,则a7/b8=
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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

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