Difference between revisions of "2003 AMC 10B Problems/Problem 8"

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Now, since <math> ar=2 </math>, <math> a=\frac{2}{r} </math>, so <math> a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3} </math>.
 
Now, since <math> ar=2 </math>, <math> a=\frac{2}{r} </math>, so <math> a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3} </math>.
  
We therefore see that <math> \boxed{\text{C}} </math> is a possible first term.
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We therefore see that <math> \boxed{\text{B}} </math> is a possible first term.
  
 
==See Also==
 
==See Also==

Revision as of 13:15, 12 November 2011

Problem

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

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

Solution

Let the first term be $a$ and the common difference be $r$. Therefore,

$ar=2$ (1)

and

$ar^3=6$. (2)

Dividing (2) by (1) eliminates the $a$, yielding $r^2=3$, so $r=\pm\sqrt{3}$.

Now, since $ar=2$, $a=\frac{2}{r}$, so $a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$.

We therefore see that $\boxed{\text{B}}$ is a possible first term.

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AMC 10 Problems and Solutions
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