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2002 AMC 12B Problem 9

Problem 9 of 25EasierAlgebra

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

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Solution

Let b=a+r,b=a+r, c=a+2r,c=a+2r, d=a+3r.d=a+3r. The geometric condition b2=adb^2=ad gives (a+r)2=a(a+3r),(a+r)^2=a(a+3r), i.e. r2=ar,r^2=ar, so r=a.r=a. Then d=a+3a=4ad=a+3a=4a and ad=14.\dfrac{a}{d}=\dfrac14. Thus, the correct answer is C.

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Concepts: arithmetic sequence · geometric sequence

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.