HMMT November

Algebra Difficulty 4.9 AIME Prove it United States

Problem:

Tyler has an infinite geometric series with sum $10$ . He increases the first term of his sequence by $4$ and swiftly changes the subsequent terms so that the common ratio remains the same, creating a new geometric series with sum $15$ . Compute the common ratio of Tyler's series.

Solution

Solution:

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

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

Solving for $r$ with $\frac{a}{1-r} = \frac{8}{1-r} = 10$ gives $r = \frac{1}{5}$ .

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