设{an}首项为a1,公差为d1,{bn}首项为b1,公差为d2
An/Bn=[na1+(n-1)nd1/2]/[nb1+(n-1)nd2/2]
=(nd1+2a1-d1)/(nd2+2b1-d2]
=n/(3n-2)
令d1=t 由2a1-d1=0 得a1=t/2
d2=3t 2b1-d2=-2t b1=t/2
a5/b5=[(t/2)+4t]/[(t/2)+12t]=9/25
设{an}首项为a1,公差为d1,{bn}首项为b1,公差为d2
An/Bn=[na1+(n-1)nd1/2]/[nb1+(n-1)nd2/2]
=(nd1+2a1-d1)/(nd2+2b1-d2]
=n/(3n-2)
令d1=t 由2a1-d1=0 得a1=t/2
d2=3t 2b1-d2=-2t b1=t/2
a5/b5=[(t/2)+4t]/[(t/2)+12t]=9/25