Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it United States

Problem:

Tyler has an infinite geometric series with sum 1010. He increases the first term of his sequence by 44 and swiftly changes the subsequent terms so that the common ratio remains the same, creating a new geometric series with sum 1515. Compute the common ratio of Tyler's series.

Solution

Solution:

Let aa and rr be the first term and common ratio of the original series, respectively. Then a1r=10\frac{a}{1-r} = 10 and a+41r=15\frac{a+4}{1-r} = 15. Dividing these equations, we get that

a+4a=1510a=8 \frac{a+4}{a} = \frac{15}{10} \Longrightarrow a = 8

Solving for rr with a1r=81r=10\frac{a}{1-r} = \frac{8}{1-r} = 10 gives r=15r = \frac{1}{5}.

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