Let the sequence {an} be an equal ratio sequence, the sum of the first n terms is Sn, and S3 = 3a3, and find the value of the common ratio Q As above

S3=3a3
That is a1 + a1q + a1q & # 178; = 3a1q & # 178;
∴1+q-2q²=0
2q²-q-1=0
(q-1)(2q+1)=0
Ψ q = 1 or q = - 1 / 2

Let Sn be the sum of the first n terms of the equal ratio sequence {an}, if S3 = 3a3, then q is the common ratio=______ ．

When the common ratio q = 1, an = A1, so S3 = 3A1 = 3a3, in line with the meaning of the problem; when the common ratio Q ≠ 1, S3 = A1 (1 − Q3) 1 − q = 3a1q2, that is 2q2-q-1 = 0, the solution can get q = − 12, or q = 1 (rounding off). In conclusion, q = 1 or − 12, so the answer is: 1 or − 12

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