The sum of the first n terms , Sn , of a progression is given by 2n^2+3n. Using the formula Tn=Sn-Sn-1,express the nth term,Tn, of the progression in terms of n.

Question

Then, how to show that the progression is an arithmetic progression and find the sum of the terms from the eighth term to the twelfth term, and also find the least value of n such that the sum of the first n terms is greater than 2000. i am quite stuck at this question for so long

Answer 1

T_n = S_n - S_n-1 = 4n+1, i.e. the terms of the sequence are 5, 9, 13, 17....

The difference between successive terms is 4.

The difference between S₁₃ and S_{7 is the sum of terms 8 through 12.}

The first n for which S_n exceeds 2000 can be fund from the solution of 2n² + 3n ≥2000