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{ | + | 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 
and 
. Which of the following is a possible first term?
Solution
Let the first term be 
and the common difference be 
. Therefore,
(1)
and
. (2)
Dividing (2) by (1) eliminates the , yielding 
, so 
.
Now, since , 
, so 
.
We therefore see that 
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 | ||
