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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 \url{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}.
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<math>a =</math> The first term in the geometric sequence.
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<math>r =</math> The ratio relating the terms of the \url{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}.
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<math>r =</math> The ratio relating the terms of the geometric sequence.
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<math>n =</math> The nth value of the \url{https://artofproblemsolving.com/wiki/index.php/Geometric_sequence}{geometric sequence}, starting at 1 and increasing as consecutive integer values.
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<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:
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~PhysicsDolphin
 
~PhysicsDolphin
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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}}

Revision as of 06:26, 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
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All AHSME Problems and Solutions

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