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Difference between revisions of "2003 AMC 10B Problems/Problem 8"
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==Solution==
 
==Solution==
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Let the first term be <math> a </math> and the common difference be <math> r </math>. Therefore,  
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Let the first term be <math> a </math> and the common ratio be <math> r </math>. Therefore,  
  
 
<cmath>ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)</cmath>
 
<cmath>ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)</cmath>
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==See Also==
 
==See Also==
  
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{{AMC10 box|year=2003|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2003|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2003|ab=B|num-b=5|num-a=7}}
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{{AMC10 box|year=2003|ab=B|num-b=7|num-a=9}}
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:44, 27 March 2023

The following problem is from both the 2003 AMC 12B #6 and 2003 AMC 10B #8, so both problems redirect to this page.

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 ratio 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 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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