It is known that the general term formula of arithmetic sequence {an} is | an | = 3n-5, and the sum formula of the first n terms is obtained We need to talk about it by category It is | an | = 16-3n

It is known that the general term formula of arithmetic sequence {an} is | an | = 3n-5, and the sum formula of the first n terms is obtained We need to talk about it by category It is | an | = 16-3n

You didn't get it right for a long time
It should be an = 16-3n
Finding the sum of the first n terms of {an}
If an = 16-3n > 0, n ≤ 5
When n ≤ 5, | an | = an
Sn=(a1+an)*n/2=(13+16-3n)*n/2
=(29-3n)n/2=29/2*n-3/2*n²
When n ≥ 6, an

In the known arithmetic sequence {an}, D ＞ 0, a3a7 = - 16, A2 + A8 = 0, let TN = | A1 | + | A2 | + +|An |. Find the general formula an of (I) {an}; (II) find TN

(1) From the properties of arithmetic sequence, we can get A2 + A8 = A3 + A7 = 0, ∵ a3a7 = - 16, and D ＞ 0 (2 points) ∵ A3 = - 4, a7 = 4, 4D = a7-a3 = 8 ∵ d = 2 ∵ an = A3 + (n-3) d = - 4 + 2 (n-3) = 2n-10 (6 points) (II) when 1 ≤ n ≤ 5, TN = | A1 | + | A2 | + +|an|=-（a1+a2+… an）=-−8+2n−102•n＝9n−n2．… When n ≥ 6, TN = | A1 | + | A2 | + +|an|=-（a1+a2+… a5）+a6+a7+… +an=-2（a1+a2+… +a5）+a1+a2+… +In conclusion: TN = 9N − N2 (1 ≤ n ≤ 5) N2 − 9N + 40 (n ≥ 6) (13 points)

In the known arithmetic sequence {an}, d > 0. A3a7 = - 16. A2 + A8 = 0. Let TN = | A1 | + | A2 | +. + | an |, (1) find {an}

First of all, A3 + A7 = A2 + A8 = 0, and because a3a7 = -16, so A3 = -4, a7 = 4, or A3 = 4, a7 = -4 (excluding because d > 0), a7-a3 = 4D = 8, d = 2, an = -4 + (n-3) * 2 = 2n-10, and because a3a7 = -16, so A3 = -4, a7 = 4, a7 = 4, or A3 = 4, a7 = - 4, a7 = - 4, or A3 = 4, a7 = - 4 = 4D = 8, d = 8, d = 8, d = 2, an = - 4 + (n-3-3) - 4 + (n-3) * (n (9-N) when n > 5, TN = T5 = T5 + T5 + T5 + A6 + A6 + A6 + A6 + A6 + A6 + A6 +... +... +...... +... And this is T5 + T5 + A6 + 20 + (A6 + an) (N-5) / 2 = 20 + (n-4) (N-5) = n ^ 2-9n + 40