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Difference between revisions of "2003 AMC 12B Problems/Problem 6"
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==Solution==
 
==Solution==
 
Call the first term <math>a_1</math> and the common ratio <math>r</math>. Then the nth term is given by <math>a_n=a_1r^{n-1}</math>. Therefore, <math>a_2=a_1r</math> and <math>a_4=a_1r^3</math>. Substituting 2 and 6 for <math>a_2</math> and <math>a_4</math> respectively gives <math>2=a_1r</math> and <math>6=a_1r^3</math>. Dividing the first equation by the second gives <math>r^2=3</math>, so <math>r=\pm\sqrt{3}</math> Substituting this into the first equation gives <math>2=a_1\pm\sqrt{3}</math>. Dividing by <math>\pm\sqrt{3}</math> and rationalizing the denominator gives <math>a_1=\pm\frac{2\sqrt{3}}{3}</math>. The negative value corresponds to answer choice <math>\boxed{\text{(B) }\frac{-2\sqrt{3}}{3}}</math>
 
Call the first term <math>a_1</math> and the common ratio <math>r</math>. Then the nth term is given by <math>a_n=a_1r^{n-1}</math>. Therefore, <math>a_2=a_1r</math> and <math>a_4=a_1r^3</math>. Substituting 2 and 6 for <math>a_2</math> and <math>a_4</math> respectively gives <math>2=a_1r</math> and <math>6=a_1r^3</math>. Dividing the first equation by the second gives <math>r^2=3</math>, so <math>r=\pm\sqrt{3}</math> Substituting this into the first equation gives <math>2=a_1\pm\sqrt{3}</math>. Dividing by <math>\pm\sqrt{3}</math> and rationalizing the denominator gives <math>a_1=\pm\frac{2\sqrt{3}}{3}</math>. The negative value corresponds to answer choice <math>\boxed{\text{(B) }\frac{-2\sqrt{3}}{3}}</math>
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{{MAA Notice}}

Revision as of 10:25, 4 July 2013

The second and fourth terms of a geometric sequence are 2 and 6. Which of the following is a possible first term?

$\text {(A) } -\sqrt{3} \qquad \text {(B) } \frac{-2\sqrt{3}}{3} \qquad \text {(C) } \frac{-\sqrt{3}}{3} \qquad \text {(D) } \sqrt{3} \qquad \text {(E) } 3$

Solution

Call the first term $a_1$ and the common ratio $r$. Then the nth term is given by $a_n=a_1r^{n-1}$. Therefore, $a_2=a_1r$ and $a_4=a_1r^3$. Substituting 2 and 6 for $a_2$ and $a_4$ respectively gives $2=a_1r$ and $6=a_1r^3$. Dividing the first equation by the second gives $r^2=3$, so $r=\pm\sqrt{3}$ Substituting this into the first equation gives $2=a_1\pm\sqrt{3}$. Dividing by $\pm\sqrt{3}$ and rationalizing the denominator gives $a_1=\pm\frac{2\sqrt{3}}{3}$. The negative value corresponds to answer choice $\boxed{\text{(B) }\frac{-2\sqrt{3}}{3}}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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