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=?
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![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=?](/uploads/image/z/13570649-17-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%2C%7Bbn%7D%2C%E5%9D%87%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%88%86%E5%88%AB%E4%B8%BASn%2CTn%2C%E8%8B%A5Sn%2FTn%3D7n%2B1%2Fn%2B3%E5%88%99a2%2Ba5%2Ba17%2Ba22%2Fb8%2Bb10%2Bb12%2Bb16%3D%3F)
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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