2013 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2013 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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Concepts:recursion work backwards

Difficulty rating: 1270

The sequence $S_1, S_2, S_3, \ldots, S_{10}$ 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 $S_4?$

$4$

$6$

$10$

$12$

$16$

Solution:

Since $S_9 = S_7 + S_8,$ we get $S_8 = 110 - 42 = 68.$ Then $S_6 = S_8 - S_7 = 68 - 42 = 26,$ $S_5 = S_7 - S_6 = 42 - 26 = 16,$ and $S_4 = S_6 - S_5 = 26 - 16 = 10.$

Thus, the correct answer is C.

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