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2013 AMC 12A Problem 7

Problem 7 of 25EasierAlgebra

The sequence S1,S_1, S2,S_2, S3,S_3, ,\ldots, S10S_{10} has the property that every term beginning with the third is the sum of the previous two. That is, Sn=Sn2+Sn1 for n3.S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. Suppose that S9=110S_9 = 110 and S7=42.S_7 = 42. What is S4?S_4?

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Solution

Since S9=S7+S8,S_9 = S_7 + S_8, we get S8=11042=68.S_8 = 110 - 42 = 68. Then S6=S8S7=6842=26,S_6 = S_8 - S_7 = 68 - 42 = 26, S5=S7S6=4226=16,S_5 = S_7 - S_6 = 42 - 26 = 16, and S4=S6S5=2616=10.S_4 = S_6 - S_5 = 26 - 16 = 10. Thus, the correct answer is C.

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