请问等差数列{An}和{Bn}的前n项和分别为Sn和Tn,且Sn/Tn=(2n+1)/(3n+2),则（a2+a5+a17+a22）/（b8+b10+b12+b16)=?顺便请问以后这种题该怎么做?

Question

请问等差数列{An}和{Bn}的前n项和分别为Sn和Tn,且Sn/Tn=(2n+1)/(3n+2),
则（a2+a5+a17+a22）/（b8+b10+b12+b16)=?
顺便请问以后这种题该怎么做?

Answer 1

a2 = a1 +d
a5 = a1 +4d
a17 = a1 + 16d
a22 = a1 + 21d
分子 = 4 a1 + 42d
b8 = b1 + 7m
b10 = b1 +9 m
b12 = b1 + 11m
b16 = b1 + 15m
分母 = 4b1 + 42m
（a2+a5+a17+a22）/（b8+b10+b12+b16)
= 2(2a1+21d)/2(2b1+21m)
= (a1+a1+21d)/(b1+b1+21m)
=(a1+a22)/(b1+b22)
= [(a1+a22)/2*22]/[(b1+b22)/2 *22]
= s22/T22
= (2*22+1)/(3*22+2)
=45/68