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

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Let the first term be <math> a </math> and the common difference be <math> r </math>. Therefore,  
 
Let the first term be <math> a </math> and the common difference be <math> r </math>. Therefore,  
  
<math> ar=2 </math> '''(1)'''
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<cmath>ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)</cmath>
  
and
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Dividing <math>(2)</math> by <math>(1)</math> eliminates the <math> a </math>, yielding <math> r^2=3 </math>, so <math> r=\pm\sqrt{3} </math>.
 
 
<math> ar^3=6 </math>. '''(2)'''
 
 
 
Dividing '''(2)''' by '''(1)''' eliminates the <math> a </math>, yielding <math> r^2=3 </math>, so <math> r=\pm\sqrt{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>.
 
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{B}} </math> is a possible first term.
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We therefore see that <math> \boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}} </math> is a possible first term.
  
 
==See Also==
 
==See Also==

Revision as of 17:59, 26 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,

\[ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)\]

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 $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ 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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