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

Revision as of 16:12, 14 July 2022

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.

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.

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 ==See Also==
+{{AMC10 box|year=2003|ab=B|num-b=7|num-a=9}}
 {{AMC12 box|year=2003|ab=B|num-b=5|num-a=7}}
-{{AMC10 box|year=2003|ab=B|num-b=7|num-a=9}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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

Revision as of 16:12, 14 July 2022

Problem

Solution

See Also