Let Sn be the sum of the first n terms of the equal ratio sequence {an}, S3, S9 and S6 be equal difference sequence, and A2 + A5 = 2am, then M is equal to () A. 6B. 7C. 8D. 10

If q = 1, then there are S3 = 3A1, S6 = 6A1, S9 = 9a1. But A1 ≠ 0, that is, S3 + S6 ≠ 2s9, which is contradictory to the problem, Q ≠1. According to S3 + S6 = 2s9, we can get A1 (1 − Q3) 1 (1 Q33) 1 − Q + A1 (6) (1 A1 (6) (1 A1 (1 − Q9) 1 − q = A1 (1 − Q9) 1 − Q (1) 1) 1 − Q (1) 1-q3 3-q3 (2q6-q3-3-1) = 0. From Q ≠ 0, we can get the equation 2q6-q36-q36-q3 6-q3-q3-1 = 0 (2q6-q3-6-q3-q3-q3-1-1 = 0. (2q3 + 1 = 0  Q3 = - 12, A2 + A5 = 2am, A2 + a2q3 = 2a2qm-2 ）M − 23, M = 8, so C

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