Skip to main content

2013 AMC 12A Problem 14

Problem 14 of 25IntermediateAlgebra

The sequence log⁡12162, log⁡12x, log⁡12y, log⁡12z, log⁡121250 \begin{gathered} \log_{12} 162, \ \log_{12} x, \ \log_{12} y, \\ \ \log_{12} z, \ \log_{12} 1250 \end{gathered} is an arithmetic progression. What is x?x?

Answer choices

Show solution

Solution

Because the logarithms are in arithmetic progression, 162,x,y,z,1250162, x, y, z, 1250 is a geometric sequence. Its common ratio rr satisfies 162r4=1250,162 r^4 = 1250, so r4=62581r^4 = \tfrac{625}{81} and r=53.r = \tfrac53. Therefore x=162⋅53=270.x = 162\cdot\tfrac53 = 270. Thus, the correct answer is B.

More practice

Concepts: logarithm · geometric sequence · arithmetic sequence

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