2007 AMC 12B Problem 15

Attempt Problem 15 of the 2007 AMC 12B 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 2007 AMC 12B solutions, or check the answer key.

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15.

The geometric series a+ar+ar2+⋯a+ar+ar^2+\cdots has a sum of 7,7, and the terms involving odd powers of rr have a sum of 3.3. What is a+r?a+r?
43\dfrac{4}{3}
127\dfrac{12}{7}
32\dfrac{3}{2}
73\dfrac{7}{3}
52\dfrac{5}{2}
Answer: E
Concepts:geometric sequence
Small hint:
The even-power terms sum to 7−3=47-3=4
Big hint:
The odd-power series is rr times the even-power series, so 3=r⋅43=r\cdot4
Solution:
The odd-power terms are ar+ar3+⋯ar+ar^3+\cdots =r(a+ar2+⋯ ),=r(a+ar^2+\cdots), that is, rr times the even-power terms. The even-power terms sum to 7−3=4.7-3=4.
So 3=4r,3=4r, giving r=34.r=\tfrac34. Then a=7(1−r)=74,a=7(1-r)=\tfrac74, and a+r=74+34=52. a+r=\dfrac74+\dfrac34=\dfrac52.
Thus, the correct answer is E.
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