Question

两个等差数列{an},(bn}前n项和分别为Sn,S'n,若Sn/S'n=(2n+3)/(3n-1)求a9/b9

Answer 1

an=a+(n-1)d

bn=b+(n-1)c

则Sn=(2a+nd-d)*n/2

S'n=(2b+cn-c)*n/2

Sn/S'n=[2a+(n-1)d]/[2b+(n-1)c]=(2n+3)/(3n-1)

a9/b9=(a+8d)/(b+8c)=(2a+16d)/(2a+16c)

则(2a+16d)/(2a+16c)就是n=17时[2a+(n-1)d]/[2b+(n-1)c]的值

n=17,[2a+(n-1)d]/[2b+(n-1)c]=(2n+3)/(3n-1)

所以a9/b9=(2*17+3)/(3*17-1)=37/50