Difference between revisions of "1975 AHSME Problems/Problem 16"
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Let's establish some ground rules... | Let's establish some ground rules... | ||
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<math>a =</math> The first term in the geometric sequence. | <math>a =</math> The first term in the geometric sequence. | ||
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<math>r =</math> The ratio relating the terms of the geometric sequence. | <math>r =</math> The ratio relating the terms of the geometric sequence. | ||
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<math>n =</math> The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values. | <math>n =</math> The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values. | ||
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Using these terms, the sum can be written as: | Using these terms, the sum can be written as: | ||
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<math>sum = a/(1-r) = 3</math> | <math>sum = a/(1-r) = 3</math> | ||
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Let <math>x =</math> The positive integer that is in the reciprocal of the geometric ratio. | Let <math>x =</math> The positive integer that is in the reciprocal of the geometric ratio. | ||
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This gives: | This gives: | ||
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<math>3 = a/(1-(1/x))</math> | <math>3 = a/(1-(1/x))</math> | ||
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<math>3 = ax/(x-1)</math> | <math>3 = ax/(x-1)</math> | ||
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Now through careful inspection we notice that when x = 3 the equation becomes | Now through careful inspection we notice that when x = 3 the equation becomes | ||
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<math>3 = a(3/2)</math>, where <math>a = 2</math>. | <math>3 = a(3/2)</math>, where <math>a = 2</math>. | ||
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Therefore <math>r = 1/x = 1/3</math>. Now we define the sum as <math>2 * (1/3)^{n-1}</math>. | Therefore <math>r = 1/x = 1/3</math>. Now we define the sum as <math>2 * (1/3)^{n-1}</math>. | ||
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Now we simply add the <math>n = 1</math> and <math>n = 2</math> terms. | Now we simply add the <math>n = 1</math> and <math>n = 2</math> terms. | ||
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<math>sum = 2(1/3)^{(1)-1} + 2(1/3)^{(2)-1} = 2 + 2/3 = 6/3 + 2/3 = 8/3</math> | <math>sum = 2(1/3)^{(1)-1} + 2(1/3)^{(2)-1} = 2 + 2/3 = 6/3 + 2/3 = 8/3</math> | ||
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This gives <math>\boxed{\textbf{(C) } 8/3}</math>. | This gives <math>\boxed{\textbf{(C) } 8/3}</math>. | ||
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~PhysicsDolphin | ~PhysicsDolphin | ||
Latest revision as of 06:27, 1 July 2024
Problem
If the first term of an infinite geometric series is a positive integer, the common ratio is the reciprocal of a positive integer, and the sum of the series is 
, then the sum of the first two terms of the series is
Solution
Let's establish some ground rules...
The first term in the geometric sequence.
The ratio relating the terms of the geometric sequence.
The nth value of the geometric sequence, starting at 1 and increasing as consecutive integer values.
Using these terms, the sum can be written as:
Let 
The positive integer that is in the reciprocal of the geometric ratio.
This gives:
Now through careful inspection we notice that when x = 3 the equation becomes
, where 
.
Therefore 
. Now we define the sum as 
.
Now we simply add the 
and 
terms.
This gives 
.
~PhysicsDolphin
For more information on geometric sequences: https://artofproblemsolving.com/wiki/index.php/Geometric_sequence
See Also
| 1975 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
