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Philippine Mathematical Olympiad

Algebra Difficulty 5.3 AIME, harder Prove it Philippines

Problem:

An infinite geometric series has sum 20202020. If the first term, the third term, and the fourth term form an arithmetic sequence, find the first term.

Solution

Solution:

Let aa be the first term and rr be the common ratio. Thus, a1−r=2020\frac{a}{1-r} = 2020, or a=2020(1−r)a = 2020(1 - r).

We also have ar2−a=ar3−ar2a r^{2} - a = a r^{3} - a r^{2}. Since the sum of the geometric series is nonzero, a≠0a \neq 0, and so we have r2−1=r3−r2r^{2} - 1 = r^{3} - r^{2}, or r3−2r2+1=0r^{3} - 2 r^{2} + 1 = 0.

Since the sum of the geometric series is finite, rr cannot be 11, so r3−2r2+1=(r−1)(r2−r−1)=0r^{3} - 2 r^{2} + 1 = (r - 1)(r^{2} - r - 1) = 0 implies r2−r−1=0r^{2} - r - 1 = 0.

Solving this equation gives r=1−52r = \frac{1 - \sqrt{5}}{2} (since ∣r∣<1|r| < 1). This gives us a=2020(1+52)=1010(1+5)a = 2020 \left( \frac{1 + \sqrt{5}}{2} \right ) = 1010(1 + \sqrt{5}).

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