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2004 AMC 12A Problem 14

Problem 14 of 25IntermediateAlgebra

A sequence of three real numbers forms an arithmetic progression with a first term of 9.9. If 22 is added to the second term and 2020 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?

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Solution

The arithmetic progression is 9,9, 9+d,9 + d, 9+2d.9 + 2d. After the additions, the geometric progression is 9,9, 11+d,11 + d, 29+2d.29 + 2d. The geometric condition gives (11+d)2=9(29+2d),(11 + d)^2 = 9(29 + 2d), which simplifies to d2+4d−140=0,d^2 + 4d - 140 = 0, so d=10d = 10 or d=−14.d = -14. The corresponding third terms 29+2d29 + 2d are 4949 and 1,1, so the smallest possible value is 1.1. Thus, the correct answer is A.
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