1952 AMC 12 Problem 12

Attempt Problem 12 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

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12.

The sum to infinity of the terms of an infinite geometric progression is

6.

The sum of the first two terms is

4\dfrac12.

The first term of the progression is:

3

1\dfrac12

1

2\dfrac12

6

9

3

Let the first term be

a

and the common ratio be

r

Use

a=6(1-r)

in the equation

a(1+r)=\frac{9}{2}

Let the first term be

a

and the common ratio be

r.

The infinite sum gives

a=6(1-r).

The first two terms therefore satisfy

a(1+r)=6(1-r^2)=\frac92.

Thus

r^2=\frac{1}{4},

r=\frac{1}{2}

gives

a=3,

while

r=-\frac{1}{2}

gives

a=9.

Thus, the correct answer is E.

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