Difference between revisions of "1975 AHSME Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | + | ||
+ | Let's establish some ground rules... | ||
+ | |||
+ | <math>a =</math> The first term in the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}. | ||
+ | <math>r =</math> The ratio relating the terms of the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}. | ||
+ | <math>n =</math> The nth value of the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}, starting at 1 and increasing as consecutive integer values. | ||
+ | |||
+ | Using these terms, the sum can be written as: | ||
+ | <math>sum = a/(1-r) = 3</math> | ||
+ | |||
+ | Let <math>x =</math> The positive integer that is in the reciprocal of the geometric ratio. | ||
+ | |||
+ | This gives: | ||
+ | <math>3 = a/(1-(1/x))</math> | ||
+ | <math>3 = ax/(x-1)</math> | ||
+ | |||
+ | Now through careful inspection we notice that when x = 3 the equation becomes | ||
+ | <math>3 = a(3/2)</math>, where <math>a = 2</math>. | ||
+ | |||
+ | Therefore <math>r = 1/x = 1/3</math>. Now we define the sum as <math>2 * ((1/3)^(n-1))</math>. | ||
+ | |||
+ | Now we simply add the <math>n = 1</math> and <math>n = 2</math> terms. | ||
+ | |||
+ | <math>sum = 2(1/3)^((1)-1) + 2(1/3)^((2)-1) = 2 + 2/3 = 6/3 + 2/3 = 8/3</math> | ||
+ | |||
+ | This gives <math>\boxed{\textbf{(C) } 8/3}</math>. | ||
+ | |||
+ | ~PhysicsDolphin | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1975|num-b=15|num-a=17}} | {{AHSME box|year=1975|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:11, 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 \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}.
The ratio relating the terms of the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}.
The nth value of the \href{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{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
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.