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an,bn是等差数列,前n项和为snT你,sn/tn=n+1/2n-3,求a5/b7
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an,bn是等差数列,前n项和为snT你,sn/tn=n+1/2n-3,求a5/b7
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答案和解析
设{an}公差为d；{bn}公差为d'.
Sn：Tn=[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)]
=(n+1)/(2n-3)
令d=t,则2a1-d=t,d'=2t,2b1-d'=-3t
解得
a1=t d=t b1=-t/2 d'=2t
a5/b7=(a1+4d)/(b1+6d')
=(t+4t)/(-t/2 +12t)
=10/23
Sn：Tn=[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)]
=(n+1)/(2n-3)
令d=t,则2a1-d=t,d'=2t,2b1-d'=-3t
解得
a1=t d=t b1=-t/2 d'=2t
a5/b7=(a1+4d)/(b1+6d')
=(t+4t)/(-t/2 +12t)
=10/23
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