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1950 AHSME Problems/Problem 5: Difference between revisions

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== Problem==
== Problem==


If five geometric means are inserted between <math>8</math> and <math>5832</math>, the fifth term in the gemetric series:
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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[[Category:Introductory Algebra Problems]]
[[Category:Introductory Algebra Problems]]
{{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
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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