### 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: To solve this problem, I need to recall what an arithmetic progression is.
An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant.
For example, \[2, 5, 8, 11, 14\] is an arithmetic progression with a common difference of 3.
So, in the given sequence, the common difference must be the same for any pair of consecutive terms.
That means that \[\log_{12}{x} - \log_{12}{162} = \log_{12}{y} - \log_{12}{x} = \log_{12}{z} - \log_{12}{y} = \log_{12}{1250} - \log_{12}{z}.\]
This equation can be simplified by using the properties of logarithms.
Specifically, I can use the fact that \[\log_{a}{b} - \log_{a}{c} = \log_{a}{\frac{b}{c}}.\]
Applying this to each term, I get \[\log_{12}{\frac{x}{162}} = \log_{12}{\frac{y}{x}} = \log_{12}{\frac{z}{y}} = \log_{12}{\frac{1250}{z}}.\]
This means that the ratios \[\frac{x}{162}, \ \frac{y}{x}, \ \frac{z}{y}, \ \frac{1250}{z}\] are all equal to some constant value.
Let's call this value $r$.
Then, I can write \[\frac{x}{162} = r, \ \frac{y}{x} = r, \ \frac{z}{y} = r, \ \frac{1250}{z} = r.\]
Now, I can solve for $x$ by using the first equation.
Multiplying both sides by 162, I get \[x = 162r.\]
To find the value of $r$, I can use any of the other equations.
For example, using the last equation, I get \[z = \frac{1250}{r}.\]
Substituting this into the third equation, I get \[\frac{z}{y} = r \implies \frac{1250}{ry} = r \implies y = \frac{1250}{r^2}.\]
Substituting this into the second equation, I get \[\frac{y}{x} = r \implies \frac{1250}{r^2x} = r \implies x = \frac{1250}{r^3}.\]
Equating this with the expression for $x$ that I found earlier, I get \[162r = \frac{1250}{r^3} \implies r^4 = \frac{1250}{162} \implies r = \sqrt[4]{\frac{1250}{162}}.\]
Therefore, the value of $x$ is \[x = 162r = 162\sqrt[4]{\frac{1250}{162}}.\]