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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Revision as of 09:25, 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?
Solution
Call the first term and the common ratio . Then the nth term is given by . Therefore, and . Substituting 2 and 6 for and respectively gives and . Dividing the first equation by the second gives , so Substituting this into the first equation gives . Dividing by and rationalizing the denominator gives . The negative value corresponds to answer choice The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.