If there are 3N items in the equal ratio sequence {an}, the sum of the first 2n items is 100 and the sum of the last 2n items is 200, then how much is the sum of the middle n items in the equal ratio sequence

If there are 3N items in the equal ratio sequence {an}, the sum of the first 2n items is 100 and the sum of the last 2n items is 200, then how much is the sum of the middle n items in the equal ratio sequence

S2n = 100, A1 (1-Q ^ 2n) / (1-Q) = 100, s3n Sn = 200, A1 (Q ^ n-q ^ 3n) / (1-Q) = 200, so Q ^ n = 2, s2n = 100 = A1 (1-Q ^ 2n) / (1-Q) = A1 (1 + Q ^ n) (1-Q ^ n) / (1-Q) = (1 + Q ^ n) Sn = 3Sn, so Sn = 100 / 3, middle n = 100-100 / 3

If the sum of the first n terms is 25 and the sum of the first 2n terms is 100, then the sum of the first 3N terms is 100______ ．

∵ the sum of the first n terms of the arithmetic sequence is 25, the sum of the first 2n terms is 100, ∵ Sn, s2n Sn, s3n-s2n become the arithmetic sequence, ∵ 25100-25, s3n-100 become the arithmetic sequence, ∵ 2 (100-25) = 25 + (s3n-100), the solution is s3n = 225

It is known that the first n terms and Sn of sequence {an} are quadratic functions of N, and A1 = - 2, A2 = 2. S3 = 6

It is solved by undetermined coefficient method,
Let Sn = BN ^ 2 + CN + D, then n = 1,2,3 be substituted into the following formula respectively
b+c+d=a1=-2
4b+2c+d=a1+a2=0
9b+3c+d=a1+a2+a3=6
By solving the equations, B = 2, C = - 4, d = 0
That is: SN = 2n ^ 2-4n