Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it

3. A quadratic sequence has three terms which are $9, 21, 37$ . Prove: 1:393 could be a term in this sequence.

Solution

3. Given $n=5, a_{m}=21, a_{k}=37$ , with common difference $d$ . We have
$\begin{array}{l} (m-n) d=21-9=12, \\ (k-m) d=37-21=16 . \\ \therefore(k-2 m+n) d=16-12=4 . \end{array}$

Let $p=k-2 m+n$ , then $p d=4$ .
$\therefore a_{k+\mu}=1993$ . Here $s=\frac{1993-37}{4}=489$ .

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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.