Difference between revisions of "1951 AHSME Problems/Problem 11"
Yankeesfan (talk | contribs) (Created page with "== 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...") |
Mathgeek2006 (talk | contribs) m |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
== 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 | + | == See Also == |
− | {{AHSME box|year=1951|num-b=10|num-a=12}} | + | {{AHSME 50p box|year=1951|num-b=10|num-a=12}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
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 | |
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.