# 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 $x=162\cdot(5/3) = 270$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 