2002 AMC 12B Problem 9

Below is the professionally curated solution for Problem 9 of the 2002 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12B solutions, or check the answer key.

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Concepts:arithmetic sequence geometric sequence

Difficulty rating: 1330

If $a,$ $b,$ $c,$ $d$ are positive real numbers such that $a,$ $b,$ $c,$ $d$ form an increasing arithmetic sequence and $a,$ $b,$ $d$ form a geometric sequence, then $\dfrac{a}{d}$ is

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Solution:

Let $b=a+r,$ $c=a+2r,$ $d=a+3r.$ The geometric condition $b^2=ad$ gives $(a+r)^2=a(a+3r),$ i.e. $r^2=ar,$ so $r=a.$

Then $d=a+3a=4a$ and $\dfrac{a}{d}=\dfrac14.$

Thus, the correct answer is C.

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