2011 AMC 10A Problem 17

Attempt Problem 17 of the 2011 AMC 10A 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 2011 AMC 10A solutions, or check the answer key.

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

17.

In the eight term sequence A,A, B,B, C,C, D,D, E,E, F,F, G,G, H,H, the value of CC is 55 and the sum of any three consecutive terms is 30.30. What is A+H?A+H?
1717
1818
2525
2626
4343
Answer: C
Concepts:recursionpattern recognition
Small hint:
Compare two consecutive three-term sums
Big hint:
The sequence repeats every three terms, so A=D=GA=D=G and C=FC=F
Solution:
From the condition about the sequence, we get that A+B+C=30 A + B + C = 30 B=25−A. B = 25 - A. Similarly, we get B+C+D=30 B + C + D = 30 D=A. D = A.
Propagating these values through the sequence and repeating the condition for every consecutive triple, we get that E=25−A,F=5,G=A, E = 25 - A, F = 5, G = A, and finally, H=25−A.H = 25 - A.
The desired sum is then A+25−A=25. A + 25 - A = 25.
Thus, C is the correct answer.
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