2013 AMC 12A Problems/Problem 14

Problem

The sequence

$\log_{12}{162}$, $\log_{12}{x}$, $\log_{12}{y}$, $\log_{12}{z}$, $\log_{12}{1250}$

is an arithmetic progression. What is $x$?

$\textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}$

Solution 1

Since the sequence is arithmetic,

$\log_{12}{162}$ + $4d$ = $\log_{12}{1250}$, where $d$ is the common difference.


Therefore,

$4d$ = $\log_{12}{1250}$ - $\log_{12}{162}$ = $\log_{12}{(1250/162)}$, and

$d$ = $\frac{1}{4}$($\log_{12}{(1250/162)}$) = $\log_{12}{(1250/162)^{1/4}}$


Now that we found $d$, we just add it to the first term to find $x$:

$\log_{12}{162}$ + $\log_{12}{(1250/162)^{1/4}}$ = $\log_{12}{((162)(1250/162)^{1/4})}$

$x$ = $(162)$$(1250/162)^{1/4}$ = $(162)$$(625/81)^{1/4}$ = $(162)(5/3)$ = $270$, which is $B$

Solution 2

As the sequence $\log_{12}{162}$, $\log_{12}{x}$, $\log_{12}{y}$, $\log_{12}{z}$, $\log_{12}{1250}$ is an arithmetic progression, the sequence $162,x,y,z,1250$ must be a geometric progression.

If we factor the two known terms we get $162=2\cdot 3^4$ and $1250=2\cdot 5^4$, thus the quotient is obviously $5/3$ and therefore $x=162\cdot(5/3) = 270$.

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions

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