1972 AMC 12 Problem 16

Attempt Problem 16 of the 1972 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 1972 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

There are two positive numbers that may be inserted between

3

and

9

such that the first three are in geometric progression while the last three are in arithmetic progression. The sum of those two positive numbers is:

13\frac12

11\frac14

10\frac12

10

9\frac12

Call the inserted numbers

x

and

y,

in that order

Use

x^2=3y

and

2y=x+9

The progression conditions give

x^2=3y

and

2y=x+9.

Eliminating

y

yields

\begin{aligned} 2x^2-3x-27&=(2x-9)(x+3)\\ &=0. \end{aligned}

Positivity gives

x=\frac{9}{2},

and then

y=\frac{27}{4}.

Their sum is

\frac92+\frac{27}{4}=\frac{45}{4}=11\frac14.

Therefore, the correct answer is B.

Problem 16 in other years