已知两个等差数列an和bn的前n项和分别为Sn,Tn ( 1)若Sn/Tn=(7n+2)/(n+3) 求an/bn 2)
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已知两个等差数列an和bn的前n项和分别为Sn,Tn ( 1)若Sn/Tn=(7n+2)/(n+3) 求an/bn 2)若an/bn
已知两个等差数列an和bn的前n项和分别为Sn,Tn (1)若Sn/Tn=(7n+2)/(n+3) 求an/bn 2)若an/bn=(14n-5)/(2n+2)求Sn/Tn
已知两个等差数列an和bn的前n项和分别为Sn,Tn (1)若Sn/Tn=(7n+2)/(n+3) 求an/bn 2)若an/bn=(14n-5)/(2n+2)求Sn/Tn
(1)
等差数列an的公差为d1
等差数列bn的公差为d2
Sn/Tn=(7n+2)/(n+3)
[n(a1+an)/2]/[n(b1+bn)/2]=(7n+2)/(n+3)
(a1+a1+(n-1)*d1)/(b1+b1+(n-1)*d2)=(7n+2)/(n+3)
(2a1-d+n*d1)/(2b1-d2+n*d2)=(7n+2)/(n+3)
于是分子分母比较相应系数
2a1-d1=2k,d1=7k,2b1-d2=3k,d2=k
解得a1=9k/2,d1=7k,b1=2k,d2=k
即
an=9k/2+(n-1)*7k)=7kn-5k/2
bn=2k+(n-1)*k)=kn+k
故
an/bn=(7kn-5k/2)/(kn+k)=(14n-5)/(2n+2)
(2)
an/bn=(14n-5)/(2n+2)
(a1+(n-1)*d1)/(b1+(n-1)*d2)=(14n-5)/(2n+2)
(a1-d1+n*d1)/(b1-d2+n*d2)=(14n-5)/(2n+2)
于是分子分母比较相应系数
a1-d1=-5k,d1=14k,b1-d2=2k,d2=2k
解得a1=9k,d1=14k,b1=4k,d2=2k
即
Sn=n*a1+n(n-1)*d1/2=9kn+7kn(n-1)
Tn=n*b1+n(n-1)*d2/2=4kn+kn(n-1)
故
Sn/Tn=[9kn+7kn(n-1)]/[4kn+kn(n-1)]=[9+7(n-1)]/[4+(n-1)]=(7n+2)/(n+3)
等差数列an的公差为d1
等差数列bn的公差为d2
Sn/Tn=(7n+2)/(n+3)
[n(a1+an)/2]/[n(b1+bn)/2]=(7n+2)/(n+3)
(a1+a1+(n-1)*d1)/(b1+b1+(n-1)*d2)=(7n+2)/(n+3)
(2a1-d+n*d1)/(2b1-d2+n*d2)=(7n+2)/(n+3)
于是分子分母比较相应系数
2a1-d1=2k,d1=7k,2b1-d2=3k,d2=k
解得a1=9k/2,d1=7k,b1=2k,d2=k
即
an=9k/2+(n-1)*7k)=7kn-5k/2
bn=2k+(n-1)*k)=kn+k
故
an/bn=(7kn-5k/2)/(kn+k)=(14n-5)/(2n+2)
(2)
an/bn=(14n-5)/(2n+2)
(a1+(n-1)*d1)/(b1+(n-1)*d2)=(14n-5)/(2n+2)
(a1-d1+n*d1)/(b1-d2+n*d2)=(14n-5)/(2n+2)
于是分子分母比较相应系数
a1-d1=-5k,d1=14k,b1-d2=2k,d2=2k
解得a1=9k,d1=14k,b1=4k,d2=2k
即
Sn=n*a1+n(n-1)*d1/2=9kn+7kn(n-1)
Tn=n*b1+n(n-1)*d2/2=4kn+kn(n-1)
故
Sn/Tn=[9kn+7kn(n-1)]/[4kn+kn(n-1)]=[9+7(n-1)]/[4+(n-1)]=(7n+2)/(n+3)
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