If x, y, z are positive real numbers such that $$x^{12} = y^{16} = z^{24}$$, and the three quantities $$3\log_{y}x, 4 \log_{z}y, n\log_{x}z$$ are in arithmetic progression, then the value of n is

Let, $$x^{12}=y^{16}=z^{24}=k$$

So, $$\log x=\dfrac{\log k}{12}$$

$$\log y=\dfrac{\log k}{16}$$

$$\log z=\dfrac{\log k}{24}$$

Now, it is given, $$3\log_{y}x, 4 \log_{z}y, n\log_{x}z$$ are in A.P.

So, $$2\cdot\left(4\log_zy\right)=3\log_yx+n\log_xz$$ ------->(1)

or, $$\dfrac{8\log y}{\log z}=\dfrac{3\log x}{\log y}+\dfrac{n\log z}{\log x}$$

Now, $$\dfrac{\log y}{\log z}=\dfrac{\dfrac{\log k}{16}}{\dfrac{\log k}{24}}=\dfrac{24}{16}=\dfrac{3}{2}$$

Also, $$\dfrac{\log x}{\log y}=\dfrac{\dfrac{\log k}{12}}{\dfrac{\log k}{16}}=\dfrac{16}{12}=\dfrac{4}{3}$$

also, $$\dfrac{\log z}{\log x}=\dfrac{\dfrac{\log k}{24}}{\dfrac{\log k}{12}}=\dfrac{12}{24}=\dfrac{1}{2}$$

So, we can rewrite equation (1) as,

$$8\left(\dfrac{3}{2}\right)=3\left(\dfrac{4}{3}\right)+n\left(\dfrac{1}{2}\right)$$

or, $$12=4+\dfrac{n}{2}$$

or, $$n=16$$