两个等差数列anbn的前n项和分别为Sn、Tn,且Sn/Tn=(n-9)/(5n+3),那么(a14+a17+a26)/(b5+b20+b32）=

Question

两个等差数列an bn 的前n项和分别为Sn、Tn,且Sn/Tn=(n-9)/(5n+3),那么(a14+a17+a26)/(b5+b20+b32）=

Answer 1

an=a1+(n-1)d1,bn=b1+(n-1)d2
(a14+a17+a26)/(b5+b20+b32)=(3a1+54d1)/(3b1+54d2)=(a1+18d1)/(b1+18d2)
且 S37/T37=[37(2a1+36d1)/2]/[37(2b1+36d2)/2]=(a1+18d1)/(b1+18d2)=28/188=7/47
故 (a14+a17+a26)/(b5+b20+b32)=7/47