Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

For example, in the sequence x1,x2,⋯ ,xn,⋯x_{1}, x_{2}, \cdots, x_{n}, \cdots, the sum of any three consecutive terms is 20, and x1=9,x12=7x_{1}=9, x_{12}=7. Find the value of x2000x_{2000}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Given that xn+1+xn+2+xn+3=20x_{n+1}+x_{n+2}+x_{n+3}=20
xn+xn+1+xn+2=20 x_{n}+x_{n+1}+x_{n+2}=20

Subtracting the two equations, we get xn+3=xn(n⩾1)x_{n+3}=x_{n}(n \geqslant 1). Therefore, {xn}\left\{x_{n}\right\} is a periodic sequence with a period of 3, so x4=x1=9,x12=x_{4}=x_{1}=9, x_{12}= x3=7x_{3}=7. Since x1+x2+x3=20x_{1}+x_{2}+x_{3}=20, we have x2=4x_{2}=4. Hence, x2000=x2=4x_{2000}=x_{2}=4

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.