Question

等差数列已知等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,且Sn/Tn=5n-9/n=3,则使an

Answer 1

∵S(2n-1) = (2n-1)(a1+a(2n-1))/2=(2n-1)an,

T(2n-1)= (2n-1)(b1+b(2n-1))/2=(2n-1)bn,

∴S(2n-1)/T(2n-1)= an/ bn,

∵Sn/Tn=(5n-9)/(n+3),

∴an/ bn=(5(2n-1)-9)/( (2n-1)+3)=(10n-14)/2n+2)

=(5n-7)/(n+1)=5-12/(n+1),

若an/ bn是正整数,则(n+1)=3,4,6,12.

此时an/ bn=1,2,3,4.

∴使an=t*bn成立的正整数t的所有可能取值是1,2,3,4.