UPSC NDA I 2016 Mathematics Solved Paper

Concept:If $x, y, z$ are such that $\log_x y$, $\log_z x$, $\log_y z$ are in GP and $x^3, y^3, z^3$ are in AP, then we use the properties of geometric progression (constant ratio) and arithmetic progression (constant difference) to find the relationship among $x, y, z$.Explanation:Given that $\log_x y$, $\log_z x$, $\log_y z$ are in GP.So, $\frac{\log_z x}{\log_x y} = \frac{\log_y z}{\log_z x}$.Using the change of base formula $\log_a b = \frac{\log_m b}{\log_m a}$, we rewrite each term with a common base $k$:$\log_x y = \frac{\log_k y}{\log_k x}$, $\log_z x = \frac{\log_k x}{\log_k z}$, $\log_y z = \frac{\log_k z}{\log_k y}$.Substitute into the GP condition:$\frac{\frac{\log_k x}{\log_k z}}{\frac{\log_k y}{\log_k x}} = \frac{\frac{\log_k z}{\log_k y}}{\frac{\log_k x}{\log_k z}}$.Simplify: $\frac{(\log_k x)^2}{(\log_k z)(\log_k y)} = \frac{(\log_k z)^2}{(\log_k y)(\log_k x)}$.Cross-multiply: $(\log_k x)^3 = (\log_k z)^3$ → $\log_k x = \log_k z$ → $x = z$. … (1)Now, $x^3, y^3, z^3$ are in AP.So, $y^3 - x^3 = z^3 - y^3$ → $2y^3 = x^3 + z^3$.Using (1), $2y^3 = x^3 + x^3 = 2x^3$ → $y^3 = x^3$ → $y = x$. … (2)From (1) and (2), $x = y = z$.When $x=y=z$, the numbers are both in AP (common difference $0$) and in GP (common ratio $1$).Answer:$x, y, z$ are in both AP and GP. Hence option 3 is correct.