1959 AMC 12 Problem 12

Attempt Problem 12 of the 1959 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1959 AMC 12 solutions, or check the answer key.

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12.

By adding the same constant to each of

20,

50,

100

a geometric progression results. The common ratio is:

\dfrac53

\dfrac43

\dfrac32

\dfrac12

\dfrac13

Let

c

be the added constant

For three consecutive geometric terms, the square of the middle term equals the product of the outer terms

The geometric-progression condition is

(50+c)^2=(20+c)(100+c).

Expanding and cancelling

c^2

gives

500=20c,

c=25.

The common ratio is

\frac{50+25}{20+25}=\frac{75}{45}=\frac53.

Thus, the correct answer is A.

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