2013 AMC 12A Problem 7

Attempt Problem 7 of the 2013 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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 $S_4?$

$4$

$6$

$10$

$12$

$16$

Answer: C

Concepts:recursion work backwards

Difficulty rating: 1270

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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