Difference between revisions of "1953 AHSME Problems/Problem 27"
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==Solution== | ==Solution== | ||
| − | Note the areas of these | + | Note the areas of these circles is <math>1\pi</math>, <math>\frac{\pi}{4}</math>, <math>\frac{\pi}{16}, \dots</math>. The sum of these areas will thus be <math>\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)</math>. We use the formula for an infinite geometric series to get the sum of the areas will be <math>\pi\left(\frac{1}{1-\frac{1}{4}}\right)</math>, or <math>\frac{4\pi}{3}</math>. <math>\boxed{D}</math> |
==See Also== | ==See Also== | ||
Latest revision as of 00:38, 4 February 2020
Problem
The radius of the first circle is 
inch, that of the second 
inch, that of the third 
inch and so on indefinitely. The sum of the areas of the circles is:
Solution
Note the areas of these circles is 
, 
, 
. The sum of these areas will thus be 
. We use the formula for an infinite geometric series to get the sum of the areas will be 
, or 
. 
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
