等差数列{an}{bn}的前几项和分别为Sn,Tn且Sn/Tn=7n+45/n-3,求a7/b7=?

Question

等差数列{an}{bn}的前几项和分别为Sn,Tn且Sn/Tn=7n+45/n-3,求a7/b7=?

Answer 1

S13/T13=(7×13+45)/(13-3)=136/10=68/5
∵S13/T13
=[13(a1+b13)/2]/[13(b1+b13)/2]
=(a1+a13)/(b1+b13)
=2a7/2b7=a7/b7=68/5
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