### 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 where each term is obtained by adding a constant to the previous term.
Let $d$ be the common difference of the arithmetic progression.
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 by using the logarithm property that $\log_b{a} - \log_b{c} = \log_b{\frac{a}{c}}$.
For example, the first equation becomes \[\log_{12}{\frac{x}{162}} = d.\]
Similarly, the other equations become \[\log_{12}{\frac{y}{x}} = d, \ \log_{12}{\frac{z}{y}} = d, \ \log_{12}{\frac{1250}{z}} = d.\]
Now, I can use another logarithm property that $\log_b{a} = \log_b{c}$ implies $a = c$.
For example, the first equation implies that \[\frac{x}{162} = 12^d.\]
Similarly, the other equations imply that \[\frac{y}{x} = 12^d, \ \frac{z}{y} = 12^d, \ \frac{1250}{z} = 12^d.\]
I can multiply these equations together to eliminate the variables $y$ and $z$.
I get \[\frac{x}{162} \cdot \frac{y}{x} \cdot \frac{z}{y} \cdot \frac{1250}{z} = 12^d \cdot 12^d \cdot 12^d \cdot 12^d.\]
This simplifies to \[\frac{1250}{162} = 12^{4d}.\]
I can take the logarithm of both sides with base 12 to get \[\log_{12}{\frac{1250}{162}} = \log_{12}{12^{4d}}.\]
I can use the logarithm property that $\log_b{b^c} = c$ to simplify the right-hand side.
I get \[\log_{12}{\frac{1250}{162}} = 4d.\]
Now, I can solve for $d$ by dividing both sides by 4.
I get \[d = \frac{1}{4} \log_{12}{\frac{1250}{162}}.\]
I can plug this value of $d$ into the equation for $x$.
I get \[\frac{x}{162} = 12^{\frac{1}{4} \log_{12}{\frac{1250}{162}}}.\]
I can multiply both sides by 162 to get \[x = 162 \cdot 12^{\frac{1}{4} \log_{12}{\frac{1250}{162}}}.\]
This is the value of $x$.