It is known that the sum of the first n terms of the equal ratio sequence {an} is Sn, A4 equals 2A3, S2 equals 6. Find the general term formula of the sequence {an}

It is known that the sum of the first n terms of the equal ratio sequence {an} is Sn, A4 equals 2A3, S2 equals 6. Find the general term formula of the sequence {an}

Because {an} is an equal ratio sequence A4 = 2A3
So the common ratio q = 2
Because S2 = 6 = A1 (1-Q ^ 2) / (1-Q) = A1 (1 + Q) = A1 * 3 = 6
So A1 = 2
So the general term formula of sequence {an} is an = 2 ^ n

In the equal ratio sequence {an}, A1 = 2, A4 = - 54, find an and the first n terms and Sn

Because A1 = 2a4 = − 54, so Q3 = - 27, so q = - 3, so an = 2 × (- 3) n-1sn = 2 [1 − (− 3) n] 1 − (− 3) = 1 − (− 3) N2

1. Let the sum of the first n terms of an be Sn, if A1 = 1, S6 = 4S3, then A4=
2. Let Q ＜ 1, the sum of the first n terms be SN. If A3 = 2, S4 = 5s2, then the general term an=

S6=4S3 6a1+15d=4(3a1+3d)6+15d=4(3+3d)d=2a4=a1+3d=7S4=5S2a1(1-q^4)/(1-q)=5a1(1-q^2)/(1-q)1+q^2=5q²=4Q＜1q=-2a3=2a1q²=2a1=1/2aN=a1q^(n-1)=1/2(-2)^(n-1)=(-2)^(n-2)

It is known that SN is the sum of the first n terms of an arithmetic sequence, and S3, S9 and S6 are arithmetic sequences
a. Find the common ratio Q, B of the sequence {an}, if A1 = 1, find the first n terms and TN of the sequence {n * a (3n-2)} (n belongs to n)

a) When q = 1 is, S3 = 3A1, S9 = 9a1, S6 = 6A1, then 18a1 = 9a1, A1 = 0 (rounding off) q is not equal to 1, S3 = A1 (1-Q ^ 3) / (1-Q), S9 = A1 (1-Q ^ 9) / (1-Q), S6 = A1 (1-Q ^ 6) / (1-Q), 2s9 = S3 + S6 has 2q ^ 9 = q ^ 3 + Q ^ 62q ^ 6 = 1 + Q ^ 3 solution, q = - (1 / 2) ^ (1 / 3) b) an = a1q ^ (n-1) a (...)