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Difference between revisions of "1952 AHSME Problems/Problem 12"

(Created page with "== Problem== The sum to infinity of the terms of an infinite geometric progression is <math> 6 </math>. The sum of the first two terms is <math> 4\frac{1}{2} </math>. The first ...")
 
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==Solution==
 
==Solution==
  
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This geometric sequence can be written as <math> a+ar+ar^2+ar^3+\cdots </math>. We are given that <math> a+ar=4\frac{1}{2} </math>. Using the [[Geometric sequence#Infinite Geometric Sequences|formula]] for the sum of an infinite geometric series, we know that <math> \frac{a}{1-r}=6 </math>. Solving for <math> r </math> in the second equation, we find that <math> r=\frac{6-a}{6} </math>. Plugging this into the first equation results in <math> a^2-12a+27=0 </math>, which can be factored as <math> (a-3)(a-9)=0 </math>. Hence, <math> a </math> equals <math> \boxed{\textbf{(E)}\ 3 \text{ or }9} </math>.
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This geometric sequence can be written as <math> a+ar+ar^2+ar^3+\cdots </math>. We are given that <math> a+ar=4\frac{1}{2} </math>. Using the [[Geometric sequence#Infinite Geometric Sequences|formula]] for the sum of an infinite geometric series, we know that <math> \frac{a}{1-r}=6 </math>. Solving for <math> r </math> in the second equation, we find that <math> r=\frac{6-a}{6} </math>. Plugging this into the first equation results in <math> a^2-12a+27=0 </math>, which can be factored as <math> (a-3)(a-9)=0 </math>. Hence, <math> a </math> equals <math> \boxed{\textbf{(E)}\ 9 \text{ or }3} </math>.
  
 
==See also==
 
==See also==
 
{{AHSME 50p box|year=1952|num-b=11|num-a=13}}
 
{{AHSME 50p box|year=1952|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:17, 17 January 2016

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 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AHSME Problems and Solutions

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