Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it Philippines

Problem:

A geometric sequence has a nonzero first term, distinct terms, and a positive common ratio. If the second, fourth, and fifth terms form an arithmetic sequence, find the common ratio of the geometric sequence.

Solution

Solution:

Let a1a_{1} be the first term and rr the common ratio of the geometric sequence. Since the second, fourth, and fifth terms form an arithmetic sequence,
a1r3a1r=a1r4a1r3r3r=r4r30=r42r3+r0=r(r32r2+1)0=r(r1)(r2r1)r=0,1,1±52 \begin{aligned} a_{1} r^{3} - a_{1} r &= a_{1} r^{4} - a_{1} r^{3} \\ r^{3} - r &= r^{4} - r^{3} \\ 0 &= r^{4} - 2 r^{3} + r \\ 0 &= r\left(r^{3} - 2 r^{2} + 1\right) \\ 0 &= r(r-1)\left(r^{2} - r - 1\right) \\ r &= 0, 1, \frac{1 \pm \sqrt{5}}{2} \end{aligned}
Since the geometric sequence has distinct terms and a positive common ratio, the only possible value of the common ratio is 1+52\frac{1+\sqrt{5}}{2}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.