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Philippine Mathematical Olympiad

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,
a1r3−a1r=a1r4−a1r3r3−r=r4−r30=r4−2r3+r0=r(r3−2r2+1)0=r(r−1)(r2−r−1)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.