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 | + | 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 | + | <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 | + | 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 where denotes the first term and denotes the common ratio. The limit of the sum of their squares is:
Solution
Let the original geometric series be . Therefore, their squares are , which is a geometric sequence with first term and common ratio . Thus, the sum is .
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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