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设等差数列anbn的前n项和分别为Sn,Tn,若Sn/Tn=n+1/2n-3,则a5/b3=
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设等差数列an bn的前n项和分别为Sn,Tn,若Sn/Tn=n+1/2n-3,则a5/b3=
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答案和解析
设等差数列a、bn的公差分别为d、k,则
n=1时,有a1/b1=-2,故a1=-2b1,
n=2时,有(2a1+d)/(2b1+k)=3,故d=10b1+3k,
n=3时,有(3a1+3d)/(3b1+3k)=4/3,得k=-4b1,有d=-2b1,
于是a5/b3=(a1+4d)/(b1+2k)=10/7
或者参照http://zhidao.baidu.com/question/512092859.html：
Sn：Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)k/2]
=[2a1+(n-1)d]/[2b1+(n-1)k]
=[dn+(2a1-d)]/[kn+(2b1-k)]
=(n+1)/(2n-3)
令d=t,则2a1-d=t,k=2t,2b1-k=-3t
解得
a1=d=t b1=-t/2 k=2t,
于是a5/b3=(a1+4d)/(b1+2k)=10/7
n=1时,有a1/b1=-2,故a1=-2b1,
n=2时,有(2a1+d)/(2b1+k)=3,故d=10b1+3k,
n=3时,有(3a1+3d)/(3b1+3k)=4/3,得k=-4b1,有d=-2b1,
于是a5/b3=(a1+4d)/(b1+2k)=10/7
或者参照http://zhidao.baidu.com/question/512092859.html：
Sn：Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)k/2]
=[2a1+(n-1)d]/[2b1+(n-1)k]
=[dn+(2a1-d)]/[kn+(2b1-k)]
=(n+1)/(2n-3)
令d=t,则2a1-d=t,k=2t,2b1-k=-3t
解得
a1=d=t b1=-t/2 k=2t,
于是a5/b3=(a1+4d)/(b1+2k)=10/7
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