Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}

Given the positive term sequence {an}, the first n terms and Sn satisfy 10sn = an2 + 5An + 6, and A1, A3, A15 are equal proportion sequence, find the general term an of the sequence {an}

∵ 10sn = an2 + 5An + 6, ① ∵ 10A1 = A12 + 5A1 + 6, the solution is A1 = 2 or A1 = 3. Also, 10sn-1 = an-12 + 5an-1 + 6 (n ≥ 2), ② from ① - ② is & nbsp; 10An = (an2-an-12) + 5 (an-an-1), that is, (an + an-1) (an-an-1-5) = 0 ∵ an + an-1 > 0, ∵ an-an-an-1 = 5 & nbsp; (n ≥ 2). When A1 = 3, A3 = 13, A15 = 73. & nbsp; When A1 = 2, A3 = 12, A15 = 72, there are & nbsp; A32 = a1a15, ∧ A1 = 2, ∧ an = 5n-3