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The ratio of the sum of the first n terms of two arithmetic sequences is 2n / (3N + 1), and the ratio of the nth term of two arithmetic sequences is calculated
According to the fact that the sum of the first n terms of the arithmetic sequence is a quadratic function of one variable with no constant term
Let the sum of the first n terms of an and BN be Sn and TN respectively
From Sn / TN = 2n / (3N + 1)
Let Sn = 2Kn & # 178; TN = kn (3N + 1)
So when n ≥ 2,
an/bn=[Sn-S(n-1)]/[Tn-T(n-1)]
=[2kn²-2k(n-1)²]/[kn(3n+1)-k(n-1)(3n-2)]
=2[n²-n²+2n-1]/[3n²+n-(3n²-2n-3n+2)]
=2(2n-1)/(6n-2)
=(2n-1)/(3n-1)
When n = 1, A1 / B1 = 2 / 4 = 1 / 2, also satisfy an / BN = (2n-1) / (3n-1)
To sum up: the ratio of the nth term of two sequences an / BN = (2n-1) / (3n-1)
Sum of the first n terms of arithmetic sequence = 48, sum of the first 2n terms = 60, sum of the first 3N terms
Then the sum of terms n + 1 to 2n is 60-48 = 12. Let the sum of terms 2n + 1 to 3N be a
So a = - 24
So the sum of the first 3N terms is 60-24 = 36
The sum of the first n terms of the arithmetic sequence is 40, and the sum of the 2n terms is 120. Find the sum of the 3N terms
The sum of the first 2n terms of arithmetic sequence is equal to the first n + (first n + nd)
∴120=40+40+nd,nd=80
The sum of the first 3N terms of the arithmetic sequence is equal to the first n + (first n + nd) + (first n + 2D) = first n × 3 + 3nd = 240
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