Difference between revisions of "1951 AHSME Problems/Problem 11"

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== Problem ==
 
== Problem ==
The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 \minus{} r}</math> where <math> a</math> denotes the first term and <math> \minus{} 1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
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The limit of the sum of an infinite number of terms in a geometric progression is <math> \frac {a}{1 - r}</math> where <math> a</math> denotes the first term and <math> - 1 < r < 1</math> denotes the common ratio. The limit of the sum of their squares is:
  
<math> \textbf{(A)}\ \frac {a^2}{(1 \minus{} r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 \plus{} r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 \minus{} r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 \plus{} r^2} \qquad\textbf{(E)}\ \text{none of these}</math>
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<math> \textbf{(A)}\ \frac {a^2}{(1 - r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 - r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 + r^2} \qquad\textbf{(E)}\ \text{none of these}</math>
  
 
== Solution ==  
 
== Solution ==  
Let the original geometric series be <math>a,ar,ar^2,ar^3,ar^4\cdots</math>. Therefore, their squares are <math>a^2,a^2r^2,a^2r^4,a^2r^6,\cdots</math>, which is a [[Geometric sequence|geometric sequence]] with first term <math>a^2</math> and common ratio <math>r^2</math>. Thus, the sum is <math>\boxed{\textbf{(C)}\ \frac {a^2}{1 \minus{} r^2}}</math>.
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Let the original geometric series be <math>a,ar,ar^2,ar^3,ar^4\cdots</math>. Therefore, their squares are <math>a^2,a^2r^2,a^2r^4,a^2r^6,\cdots</math>, which is a [[Geometric sequence|geometric sequence]] with first term <math>a^2</math> and common ratio <math>r^2</math>. Thus, the sum is <math>\boxed{\textbf{(C)}\ \frac {a^2}{1 - r^2}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:56, 13 March 2015

Problem

The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 - r}$ where $a$ denotes the first term and $- 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is:

$\textbf{(A)}\ \frac {a^2}{(1 - r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 - r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 + r^2} \qquad\textbf{(E)}\ \text{none of these}$

Solution

Let the original geometric series be $a,ar,ar^2,ar^3,ar^4\cdots$. Therefore, their squares are $a^2,a^2r^2,a^2r^4,a^2r^6,\cdots$, which is a geometric sequence with first term $a^2$ and common ratio $r^2$. Thus, the sum is $\boxed{\textbf{(C)}\ \frac {a^2}{1 - r^2}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AHSME Problems and Solutions

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