1952 AHSME Problems/Problem 12

Problem

The sum to infinity of the terms of an infinite geometric progression is $6$ . The sum of the first two terms is $4\frac{1}{2}$ . The first term of the progression is:

$\textbf{(A) \ }3 \text{ or } 1\frac{1}{2} \qquad \textbf{(B) \ }1 \qquad \textbf{(C) \ }2\frac{1}{2} \qquad \textbf{(D) \ }6 \qquad \textbf{(E) \ }9\text{ or }3$

Solution

This geometric sequence can be written as $a+ar+ar^2+ar^3+\cdots$ . We are given that $a+ar=4\frac{1}{2}$ . Using the formula for the sum of an infinite geometric series, we know that $\frac{a}{1-r}=6$ . Solving for $r$ in the second equation, we find that $r=\frac{6-a}{6}$ . Plugging this into the first equation results in $a^2-12a+27=0$ , which can be factored as $(a-3)(a-9)=0$ . Hence, $a$ equals $\boxed{\textbf{(E)}\ 3 \text{ or }9}$ .

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1952 AHSME Problems/Problem 12

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