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Difference between revisions of "1950 AHSME Problems/Problem 5"

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== Problem==
 
== Problem==
  
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If five geometric means are inserted between <math>8</math> and <math>5832</math>, the fifth term in the gemetric series:
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If five geometric means are inserted between <math>8</math> and <math>5832</math>, the fifth term in the geometric series:
  
 
<math> \textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these} </math>
 
<math> \textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these} </math>
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==See Also==
 
==See Also==
  
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{{AHSME box|year=1950|num-b=4|num-a=6}}
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{{AHSME 50p box|year=1950|num-b=4|num-a=6}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 18:50, 13 July 2014

Problem

If five geometric means are inserted between $8$ and $5832$, the fifth term in the geometric series:

$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$

Solution

We can let the common ratio of the geometric sequence be $r$. $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$. By the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$ We divide by eight to find: $r^6=729$ $r=\pm 3$

Because $a_2$ will not be between $8$ and $5832$ if $r=-3$ we can discard it as an extraneous solution. We find $r=3$ and $a_5=a_1r^4=(8)(3)^4=\boxed{\textbf{(A)}\ 648}$

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AHSME Problems and Solutions

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