Difference between revisions of "2013 AMC 12A Problems/Problem 7"
Epicwisdom (talk | contribs) (Created page with "<math>S_9 = 110</math>, <math>S_7 = 42</math> <math>S_8 = S_9 - S_ 7 = 110 - 42 = 68</math> <math>S_6 = S_8 - S_7 = 68 - 42 = 26</math> <math>S_5 = S_7 - S_6 = 42 - 26 = 16</m...") |
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+ | == Problem== | ||
+ | |||
+ | The sequence <math>S_1, S_2, S_3, \cdots, S_{10}</math> has the property that every term beginning with the third is the sum of the previous two. That is, <cmath> S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. </cmath> Suppose that <math>S_9 = 110</math> and <math>S_7 = 42</math>. What is <math>S_4</math>? | ||
+ | |||
+ | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad </math> | ||
+ | |||
+ | ==Solution== | ||
<math>S_9 = 110</math>, <math>S_7 = 42</math> | <math>S_9 = 110</math>, <math>S_7 = 42</math> | ||
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<math>S_4 = S_6 - S_5 = 26 - 16 = 10</math> | <math>S_4 = S_6 - S_5 = 26 - 16 = 10</math> | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=6|num-a=8}} |
Revision as of 18:35, 22 February 2013
Problem
The sequence has the property that every term beginning with the third is the sum of the previous two. That is, Suppose that and . What is ?
Solution
,
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 |