1952 AHSME Problems/Problem 12

Revision as of 09:17, 17 January 2016 by Ilikepie2003 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{\textbf{(E)}\ 9 \text{ or }3}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png