S
由题意可得 S14 T14=
14(a1+a14) 2
14(b1+b14) 2= 2a7 2b7= a7 b7= 3×14+2 4×14−5= 44 51, 故答案为: 44 51.
若两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且满足S
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已知两个等差数列{an},{bn}的前n项的和分别为Sn,Tn,且S
已知等差数列{an},{bn}的前n项和分别为Sn和Tn,若S
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两等差数列{an}和{bn},前n项和分别为Sn,Tn,且S
两个等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn=2n/3n+1,求an/bn.
1.已知两个等差数列An,Bn,前n项和分别为Sn,Tn,且Sn/Tn=(2n+2)/(n+2),则An/Bn=
若两个等差数列an和bn的前n项和分别为Sn和Tn Sn/Tn=7n+3/n+3
设两个等差数列{an},{bn}的前n项和分别为Sn,Tn.若Sn/Tn=7n+1/4n+27,则a7/b7=
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