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
8 years, 3 months ago

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Comments

Note that we have b22=b1b3b_2^2=b_1b_3. Substituting bi=4ai2b_i=4a_i^2 for i=1,2,3i=1,2,3 gives (4a22)2=(4a12)(4a32)    (a22)2=(a12)(a32)(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 a22=a1a3a_2^2=a_1a_3, so a1,a2,a3a_1,a_2,a_3 form a geometric sequence as well. (NOTE: We don't have to worry about negatives since a22a_2^2 is always nonnegative and a1<a2<a3a_1<a_2<a_3.) Therefore, if a1,a2,a3a_1,a_2,a_3 form both an arithmetic sequence and a geometric sequence, then we must have a1=a2=a3a_1=a_2=a_3. The conclusion follows.

David Altizio - 8 years, 3 months ago

oh, i forgot, the three points are distinct

Jason Carlo Carranceja - 8 years, 3 months ago

a1=a2=a3

Akbarali Surani - 8 years, 3 months ago
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