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

Video Solution by OmegaLearn

https://youtu.be/RdIIEhsbZKw?t=944

~ pi_is_3.14

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

Page

Toolbox

Search

2013 AMC 12A Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Video Solution by OmegaLearn

See also