Sequences in Function

Hello! I encountered this puzzling question awhile back. However, I still could not verify if my answer is correct. The question goes like this: Points \((a_{1}, b_{1})\), \((a_{2}, b_{2})\), and \((a_{3}, b_{3})\) are distinct points that lie on the graph of \(y=4x^{2}\). \(a_{1}\), \(a_{2}\), and \(a_{3}\) form an arithmetic sequence while \(b_{1}\), \(b_{2}\), and \(b_{3}\) form a geometric sequence. Find all the possible common differences and common ratios of both sequences.

Note by Jason Carlo Carranceja
5 years, 6 months ago

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oh, i forgot, the three points are distinct

Jason Carlo Carranceja - 5 years, 6 months ago

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Note that we have \(b_2^2=b_1b_3\). Substituting \(b_i=4a_i^2\) for \(i=1,2,3\) gives \((4a_2^2)^2=(4a_1^2)(4a_3^2)\implies (a_2^2)^2=(a_1^2)(a_3^2)\). Taking the square root of both sides gives \(a_2^2=a_1a_3\), so \(a_1,a_2,a_3\) form a geometric sequence as well. (NOTE: We don't have to worry about negatives since \(a_2^2\) is always nonnegative and \(a_1<a_2<a_3\).) Therefore, if \(a_1,a_2,a_3\) form both an arithmetic sequence and a geometric sequence, then we must have \(a_1=a_2=a_3\). The conclusion follows.

David Altizio - 5 years, 6 months ago

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a1=a2=a3

Akbarali Surani - 5 years, 6 months ago

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