Difference between revisions of "2013 AMC 12A Problems/Problem 7"

Revision as of 14:56, 3 July 2013

Problem

The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, $S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.$ Suppose that $S_9 = 110$ and $S_7 = 42$ . What is $S_4$ ?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad$

Solution

$S_9 = 110$ , $S_7 = 42$

$S_8 = S_9 - S_ 7 = 110 - 42 = 68$

$S_6 = S_8 - S_7 = 68 - 42 = 26$

$S_5 = S_7 - S_6 = 42 - 26 = 16$

$S_4 = S_6 - S_5 = 26 - 16 = 10$

2013 AMC 12A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC12 box|year=2013|ab=A|num-b=6|num-a=8}}
+{{MAA Notice}}

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Difference between revisions of "2013 AMC 12A Problems/Problem 7"

Revision as of 14:56, 3 July 2013

Problem

Solution

See also