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China Mathematical Competition

Algebra Difficulty 4.8 AIME Prove it China

Suppose that arithmetic sequence $\{a_n\}$ satisfies $a_{2021} = a_{20} + a_{21} = 1$ . Then the value of $a_1$ is ______.

Solution

Let the common difference of $\{a_n\}$ be $d$ . By the given condition, it follows that
$\begin{cases} a_1 + 2020d = 1, \\ 2a_1 + 39d = 1. \end{cases}$
The solution is $a_1 = \frac{1981}{4001}$ .
$\square$

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