If \(a^x=b^y=c^z\) and \(b^2=ac\) then \(y\) equals? a) \(\frac{2xz}{(x+z)}\). b) \(\frac{xz}{2(x-z)}\). c) \(\frac{xz}{2(z-x)}\). d) \(\frac{2xz}{(x-z)}\)

Answer: Option (a) \(\frac{2xz}{x+z}\).

Solution:

Given: \(a^x = b^y = c^z\) and \(b^2 = ac\).

Let \(a^x = b^y = c^z = k\). Then,

\(a^x = k \implies a = k^{\frac{1}{x}}\)     ...(1)
\(b^y = k \implies  b= k^{\frac{1}{y}}\)    ...(2)
\(c^z = k \implies c= k^{\frac{1}{z}} \)  ...(3)

Substituting the values of \(a\), \(b\), and \(c\) from equations \((1)\), \((2)\), and \((3)\) into \(b^2 = ac\), we get:

\(\left(k^{\frac{1}{y}}\right)^2 = k^{\frac{1}{x}} \times k^{\frac{1}{z}}\)
\(\implies k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}}\)
\(\implies \frac{2}{y} = \frac{1}{x}+\frac{1}{z}\)

\(\implies \frac{2}{y} = \frac{z+x}{xz}.\)

Taking the reciprocal of both sides, we get:

\(\frac{y}{2} = \frac{xz}{x+z}\)
\(\implies y = \frac{2xz}{x+z}.\)

Therefore, option (a) \( \mathbf{\frac{2xz}{x+z}}\) is the correct answer.