If the tolerance of arithmetic sequence D ≠ 0, and A1, A2 are two of the equations X & # 178; - a3x + A4 = 0 about X, then the general term formula an of {an}= According to Weida's theorem, there is a1 + A2 = A3, A1A2 = A4, so how does a1 + A1 + D = a1 + 2D calculate?

If the tolerance of arithmetic sequence D ≠ 0, and A1, A2 are two of the equations X & # 178; - a3x + A4 = 0 about X, then the general term formula an of {an}= According to Weida's theorem, there is a1 + A2 = A3, A1A2 = A4, so how does a1 + A1 + D = a1 + 2D calculate?


Let the tolerance be d
A1 + A2 = A3 because A2 = a1 + D, A3 = a1 + 2D, so a1 + A1 + D = a1 + 2D
A1 * A2 = A4 can be obtained: A1 * (a1 + D) = a1 + 3D
D = 0 (rounding) or D = 2, A1 = 2
So the general formula is an = 2 + 2 (n-1) = 2n

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