|
|
| Line 1: |
Line 1: |
| − | The second and fourth terms of a geometric sequence are 2 and 6. Which of the following is a possible first term?
| + | #REDIRECT [[2003 AMC 10B Problems/Problem 8]] |
| − | | |
| − | <math>
| |
| − | \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
| |
| − | </math>
| |
| − | | |
| − | ==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>
| |
| − | {{MAA Notice}}
| |
