anbn是两个等差数列sntn分别是前n项和sn/tn=3n-1/2n+3则a5/b6=

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a(n) = a + (n-1)d,s(n)= na + n(n-1)d/2.
b(n) = b + (n-1)c.t(n) = nb + n(n-1)c/2.
(3n-1)/(2n+3) = s(n)/t(n) = [na + n(n-1)d/2]/[nb + n(n-1)c/2] = [2a + (n-1)d]/[2b + (n-1)c]
(3n-1)[2b+(n-1)c] = (2n+3)[2a + (n-1)d],
3n(2b-c) + 3cn^2 - (2b-c) - nc = 2n(2a-d) + 2dn^2 + 3(2a-d) + 3nd,
0 = (3c - 2d)n^2 + (6b - 3c - c - 4a+2d - 3d)n - (2b-c) - 3(2a-d),
= (3c-2d)n^2 + (6b - 4c - 4a - d)n - (2b - c + 6a - 3d)
0 = 3c-2d.d = 3c/2.
0 = 6b - 4c - 4a - d = 6b - 4c - 4a - 3c/2 = 6b - 4a - 11c/2.
0 = 2b - c + 6a - 3d = 2b - c + 6a - 9c/2 = 2b + 6a - 11c/2.
0 = 4b - 10a,b = 5a/2.
0 = 2b + 6a - 11c/2 = 5a + 6a - 11c/2 = 11a - 11c/2,a = c/2,
b = 5a/2 = 5c/4.
a(5)/b(6) = [ a + 4d] /[b + 5c] = [c/2 + 6c]/[5c/4 + 5c] = (2 + 24)/(5+20) = 26/25

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