已知数列{an},{bn},均为等差数列,前n项和分别为Sn,Tn,若Sn/Tn=7n+1/n+3则a2+a5+a17+a22/b8+b10+b12+b16=?

已知数列{an},{bn},均为等差数列,前n项和分别为Sn,Tn,若Sn/Tn=7n+1/n+3则a2+a5+a17+a22/b8+b10+b12+b16=?
已知数列{an},{bn},均为等差数列,前n项和分别为Sn,Tn,若Sn/Tn=7n+1/n+3则a2+a5+a17+a22/b8+b10+b12+b16=?

已知数列{an},{bn},均为等差数列,前n项和分别为Sn,Tn,若Sn/Tn=7n+1/n+3则a2+a5+a17+a22/b8+b10+b12+b16=?
A2+A5+A17+A22=(A1+d1)+(A1+4d1)+(A1+16d1)+(A1+21d1)=4A1+42d1=2(A1+A22)
B8+B10+B12+B16=(B1+7d2)+(B1+9d2)+(B1+11d2)+(B1+15d2)=4B1+42d2=2(B1+B22)
S22=(A1+A22)×22/2=11(A1+A22)
T22=(B1+B22)×22/2=11(B1+B22)
(A2+A5+A17+A22)/(B8+B10+B12+B16)
=2(A1+A22)/2(B1+B22)
=11(A1+A22)/11(B1+B22)
=S22/T22
=(7×22+1)/(22+3)
=31/5

fgdg

S1/T1=2=a1/b1,所以a1=2b1
S2/T2=a1+a1+d/b1+b1+c=3,4b1+d=6b1+3c d=2b1+3c
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)c/2]=[2nb1+n(n-1)(2b1+3c)/2]/[nb1+n(n-1)c/2]
=[n*(n+1)b1+3n(n-1)c/2[nb1+n(n-1)c/2]

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S1/T1=2=a1/b1,所以a1=2b1
S2/T2=a1+a1+d/b1+b1+c=3,4b1+d=6b1+3c d=2b1+3c
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)c/2]=[2nb1+n(n-1)(2b1+3c)/2]/[nb1+n(n-1)c/2]
=[n*(n+1)b1+3n(n-1)c/2[nb1+n(n-1)c/2]
n=3时
12b1+9c/3b1+3c=11/3 36b1+27c=33b1+33c
b1=2c
所以 a1=4c d=7c
A=a2+a5+a17+a22=2*a12
B=b8+b10+b12+b16=2*b9+2*b14=b1+b21+b1+b23
B/A=b11/2*a12+b12/2*a12=11b1+10c/2*(12a1+11d)+12b1+11c/2*(12a1+11d)
=22c+10c/2*(48c+77c)+24c+11c/2*(48c+77c)=67/250
答案是250/67

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