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Problem 661

AIME late

Algebra Difficulty 5.2 Find the answer

6.25 Three numbers form a geometric progression. If the second number is increased by 2, the progression becomes arithmetic, and if the last number is then increased by 9, the progression becomes geometric again. Find these numbers.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

6.25 According to the condition, the numbers $a, a q, a q^{2}$ form a geometric progression, the numbers $a, a q+2, a q^{2}$ form an arithmetic progression, and the numbers $a, a q+2, a q^{2}+9$ form a geometric progression again. Using formulas (6.4) and (6.8), we obtain the system of equations

$\left\{\begin{array}{l} a q+2=\frac{a+a q^{2}}{2} \\ (a q+2)^{2}=a\left(a q^{2}+9\right) \end{array}\right.$

from which we find two solutions: 1) $a=4, q=2$ ;
2) $a=\frac{4}{25}, q=-4$ .

Answer: $4 ; 8 ; 16$ or $\frac{4}{25} ;-\frac{16}{25} ; \frac{64}{25}$ .

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