If the \(4^{th}\), \(10^{th}\) and \(16^{th}\) terms of a GP are \(x,\,y\) and \(z\) respectively, prove that \(x,\,y,\,z\) are in GP.

Let the GP have first term \(a\) and common ratio \(r\).
\(x = T_4 = ar^3\)
\(y = T_{10} = ar^9\)
\(z = T_{16} = ar^{15}\)

For three numbers to be in GP,
\(y^2 = xz\)
Here,
\(y^2 = (ar^9)^2 = a^2r^{18}\)
\(xz = (ar^3)(ar^{15}) = a^2r^{18}\)
Hence \(y^2 = xz\)
Therefore, \(x,y,\) and \(z\) are in GP