1972 AMC 12 第 16 题

Question

在

3

与

9

之间插入两个正数，使前三个数成等比数列，后三个数成等差数列。这两个正数之和为：

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

依次把插入的两个数记为

x

和

y

Call the inserted numbers

x

and

y,

in that order

使用

x^2=3y

和

2y=x+9

Use

x^2=3y

and

2y=x+9

由数列条件得

x^2=3y

和

2y=x+9

。消去

y

，得

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

由正数条件得

x=\frac{9}{2}

，进而

y=\frac{27}{4}

。两数之和为

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

因此，正确答案是 B。

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.

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