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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 \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>a =</math> The first term in the 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.
+
 
 +
 
 +
<math>r =</math> The ratio relating the terms of the geometric sequence.
 +
 
 +
 
 +
<math>n =</math> 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:
 
Using these terms, the sum can be written as:
 +
 
<math>sum = a/(1-r) = 3</math>
 
<math>sum = a/(1-r) = 3</math>
 +
  
 
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.
 +
  
 
This gives:
 
This gives:
 +
 
<math>3 = a/(1-(1/x))</math>
 
<math>3 = a/(1-(1/x))</math>
 +
 
<math>3 = ax/(x-1)</math>
 
<math>3 = ax/(x-1)</math>
  
−
Now through trial and error we notice that when x = 3 this gives
+
 
 +
Now through careful inspection we notice that when x = 3 the equation becomes
 +
 
 
<math>3 =  a(3/2)</math>, where <math>a = 2</math>.
 
<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>.
 
  
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Now we simply add the <math>n = 1 and n = 2</math> terms.
+
Therefore <math>r = 1/x = 1/3</math>. Now we define the sum as <math>2 * (1/3)^{n-1}</math>.
  
−
<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{(B) } 8/3}</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
 
~PhysicsDolphin
 +
 +
For more information on geometric sequences: https://artofproblemsolving.com/wiki/index.php/Geometric_sequence
  
 
==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}}

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 $3$, then the sum of the first two terms of the series is

$\textbf{(A)}\ \frac{1}{3} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{8}{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \frac{9}{2} \qquad$

Solution

Let's establish some ground rules...


$a =$ The first term in the geometric sequence.


$r =$ The ratio relating the terms of the geometric sequence.


$n =$ 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:

$sum = a/(1-r) = 3$


Let $x =$ The positive integer that is in the reciprocal of the geometric ratio.


This gives:

$3 = a/(1-(1/x))$

$3 = ax/(x-1)$


Now through careful inspection we notice that when x = 3 the equation becomes

$3 = a(3/2)$, where $a = 2$.


Therefore $r = 1/x = 1/3$. Now we define the sum as $2 * (1/3)^{n-1}$.


Now we simply add the $n = 1$ and $n = 2$ terms.


$sum = 2(1/3)^{(1)-1} + 2(1/3)^{(2)-1} = 2 + 2/3 = 6/3 + 2/3 = 8/3$


This gives $\boxed{\textbf{(C) } 8/3}$.


~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

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