2007 AMC 12B Problem 15 — Solution | LIVE by Po-Shen Loh

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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 $a+ar+ar^2+\cdots$ has a sum of $7,$ and the terms involving odd powers of $r$ have a sum of $3.$ What is $a+r?$

$\dfrac{4}{3}$

$\dfrac{12}{7}$

$\dfrac{3}{2}$

$\dfrac{7}{3}$

$\dfrac{5}{2}$

Answer: E

Concepts:geometric sequence

Difficulty rating: 1580

Solution:

The odd-power terms are $ar+ar^3+\cdots$ $=r(a+ar^2+\cdots),$ that is, $r$ times the even-power terms. The even-power terms sum to $7-3=4.$

So $3=4r,$ giving $r=\tfrac34.$ Then $a=7(1-r)=\tfrac74,$ and $a+r=\dfrac74+\dfrac34=\dfrac52.$

Thus, the correct answer is E.

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