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It is known that the sum of the first n terms of two arithmetic sequences {an}, {BN} is Sn and TN respectively. If Sn / TN = (5N + 3) / (2n-1), then A9 / B9 is obtained Such as the title
S17=(A1+A17)×17/2=(A1+A1+16d)×17/2=(A1+8d)×17=17A9
Similarly, T17 = 17b9
A9/B9=S17/T17=(5×17+3)/(2×17-1)=88/33=8/3
The sum of the first n terms of two arithmetic sequences {an} and {BN} is Sn and TN respectively, Sn / TN = 2n + 3 / 3n-1, and A9 / B9 is obtained
The tolerances of {an} and {BN} are set to D1 and D2 respectively
Sn=na1+n(n-1)d1/2
Tn=nb1+n(n-1)d2/2
Sn/Tn=[2a1+(n-1)d1]/[2b1+(n-1)d2]=(2n+3)/(3n-1) （1）
Let n = 17 be substituted into (1)
(2a1+16d1)/(2a2+16d2)=(a1+8d1)/(b1+8d2)=a9/b9=37/50
a9/b9=37/50
Arithmetic sequence: two arithmetic sequences (an) and (BN) are known, and the sum of their first n terms is Sn, Sn ', respectively. If Sn / Sn' = 2n + 3 / 3n-1, A9 / B9 can be obtained
The tolerance of {an} and {BN} are set as D1 and d2sn = Na1 + n (n-1) D1 / 2Sn '= NB1 + n (n-1) D2 / 2Sn / Sn' = [2A1 + (n-1) D1] / [2B1 + (n-1) D2] = (2n + 3) / (3n-1) (1) let n = 17 be substituted into (1) formula: (2A1 + 16d1) / (2A2 + 16d2) = (a1 + 8d1) / (B1 + 8d2) = A9 / B9 = 37 / 50a9 / B9 = 37 / 50
An and BN are arithmetic sequences, the sum of the first n terms is Sn, TN respectively, and TN of Sn is equal to 3N + 1 / 2n-3, then what is A9 / B9 equal to
Tn/Sn=(3n+1)/(2n+3)=[(b1+bn)*n/2]/[(a1+an)*n/2)]
a9/b9=(2*a9)/(2*b9)=[(a1+a17)*17/2]/[(b1+b17)/*17/2]=S17/T17
=
37/52
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