Let the sum of the first n terms of the sequence {an} be Sn = n ^ 2-8n. If TN is the sum of the first n terms of the sequence {an's absolute value}, find TN

Let the sum of the first n terms of the sequence {an} be Sn = n ^ 2-8n. If TN is the sum of the first n terms of the sequence {an's absolute value}, find TN

Sn - sn-1 = n ^ 2-8n - [(n-1) ^ 2-8 (n-1)] = 2n-9 = an an an is the arithmetic sequence with tolerance 2
an

In the arithmetic sequence {an}, if the sum of the first n terms is Sn, S5 = 17, S10 = 68
The ratio of tolerance to first term is?

a3=S5/5=17/5
a8=(S10-S5)/5=51/5
a8-a3=5d=34/5
Then the tolerance d = 34 / 25
First term a = a3-2d = 17 / 5-2 × 34 / 25 = 17 / 25
Then a: D = 1:2

If the sum of the first n terms in the arithmetic sequence {an}, {BN} is an and BN respectively, and an / BN = n / 3n-2, then what is A5 / B5

Let the first term of {an} be A1, the tolerance be D1, the first term of {BN} be B1, the tolerance be d2an / BN = [Na1 + (n-1) ND1 / 2] / [NB1 + (n-1) ND2 / 2] = (ND1 + 2a1-d1) / (ND2 + 2b1-d2] = n / (3n-2) let D1 = t get A1 = t / 2d2 = 3T 2b1-d2 = - 2T B1 = t / 2a5 / B5 = [(T / 2) + 4T] / [(T / 2) + 12t