Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME Prove it Bulgaria

The first, seventh, and seventeenth terms of an arithmetic progression are distinct and consecutive terms of a geometric progression. To find the difference of the arithmetic progression if its first term is a solution of the equation
$x^2 - 9x + x\sqrt{12-x} - 9\sqrt{12-x} = 0.$

Solutions — 2

Solution 1

Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$ , $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.
$(a_1 + 6d)^2 = a_1 \cdot (a_1 + 16d) \iff d \cdot (a_1 - 9d) = 0.$
Since $d \neq 0$ , we get that $a_1 = 9d$ . Furthermore, we have $(x-9)(x+\sqrt{12-x}) = 0$ and $x \le 12$ . Then $x = 9$ or $\sqrt{12-x} = -x$ , i.e. $x^2 + x - 12 = 0$ and $x \le 0$ , whence $x = -4$ . Then $a_1 = 9$ and $a_1 = -4$ , as $d = 1$ and $d = -\frac{4}{9}$ , respectively. $\square$

Solution 2

Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$ , $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.
$(a_1 + 6d)^2 = a_1 \cdot (a_1 + 16d) \iff d \cdot (a_1 - 9d) = 0.$
Since $d \ne 0$ , we get that $a_1 = 9d$ . Furthermore, we have $(x-9)(x+\sqrt{12-x}) = 0$ and $x \le 12$ . Then $x = 9$ or $\sqrt{12-x} = -x$ , i.e. $x^2 + x - 12 = 0$ and $x \le 0$ , whence $x = -4$ . Then $a_1 = 9$ and $a_1 = -4$ , as $d = 1$ and $d = -\frac{4}{9}$ , respectively. $\square$

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