### Instruction: The sequence \[\log_{12}{162}, \ \log_{12}{x}, \ \log_{12}{y}, \ \log_{12}{z}, \ \log_{12}{1250}\]is an arithmetic progression. What is $x$?### Response: An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant.
Let $d$ be the common difference of this sequence.
Then I can write the following equations:

\[\log_{12}{x} - \log_{12}{162} = d\]
\[\log_{12}{y} - \log_{12}{x} = d\]
\[\log_{12}{z} - \log_{12}{y} = d\]
\[\log_{12}{1250} - \log_{12}{z} = d\]
I can simplify these equations using the properties of logarithms.
For example, $\log_{12}{x} - \log_{12}{162} = \log_{12}{\frac{x}{162}}$.
So the equations become:

\[\log_{12}{\frac{x}{162}} = d\]
\[\log_{12}{\frac{y}{x}} = d\]
\[\log_{12}{\frac{z}{y}} = d\]
\[\log_{12}{\frac{1250}{z}} = d\]
Now I can use the fact that $d$ is constant and eliminate it from the equations.
For example, by equating the first and second equations, I get:

\[\log_{12}{\frac{x}{162}} = \log_{12}{\frac{y}{x}}\]
This implies that $\frac{x}{162} = \frac{y}{x}$, since logarithms are one-to-one functions.
Similarly, I can eliminate $d$ from the other equations and get:

\[\frac{x}{162} = \frac{y}{x}\]
\[\frac{y}{x} = \frac{z}{y}\]
\[\frac{z}{y} = \frac{1250}{z}\]
These equations are equivalent to:

\[x^2 = 162y\]
\[y^2 = xz\]
\[z^2 = 1250y\]
I can use these equations to eliminate $y$ and $z$ in terms of $x$.
For example, from the first equation, I get $y = \frac{x^2}{162}$.
Substituting this into the second equation, I get:

\[\frac{x^4}{162^2} = xz\]
Solving for $z$, I get $z = \frac{x^3}{162^2}$.
Substituting this into the third equation, I get:

\[\frac{x^6}{162^4} = 1250 \cdot \frac{x^2}{162}\]
Simplifying this equation, I get:

\[x^4 = 1250 \cdot 162^3\]