Difference between revisions of "2003 AMC 10B Problems/Problem 8"
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+ | {{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #6]] and [[2003 AMC 10B Problems|2003 AMC 10B #8]]}} | ||
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==Problem== | ==Problem== | ||
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==Solution== | ==Solution== | ||
− | Let the first term be <math> a </math> and the common | + | 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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+ | {{AMC12 box|year=2003|ab=B|num-b=5|num-a=7}} | ||
{{AMC10 box|year=2003|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2003|ab=B|num-b=7|num-a=9}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:07, 26 March 2021
- 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 and . Which of the following is a possible first term?
Solution
Let the first term be and the common ratio be . Therefore,
Dividing by eliminates the , yielding , so .
Now, since , , so .
We therefore see that is a possible first term.
See Also
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.