2010 AMC 12B Problems/Problem 20
Contents
Problem
A geometric sequence has , , and for some real number . For what value of does ?
Solution
By the defintion of a geometric sequence, we have . Since , we can rewrite this as .
The common ratio of the sequence is , so we can write
Since , we have , which is , making our answer .
Solution 2
Notice that the common ratio is ; multiplying it to gives . Then, working backwards we have , and . Now notice that since and , we need , so . Dividing both sides by gives , which the left side is equal to ; we see as well that the right hand side is equal to given , so the answer is . - mathleticguyyy
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
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