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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. And 10sn-1 = an-12 + 5an-1 + 6 (n ≥ 2), ② from ① - ②, we get & 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 ∵
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. And 10sn-1 = an-12 + 5an-1 + 6 (n ≥ 2), ② from ① - ②, we get & 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 ∵
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. And 10sn-1 = an-12 + 5an-1 + 6 (n ≥ 2), ② from ① - ②, we get & 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 ∵
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