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 ...