UPSC NDA I 2021 Mathematics Solved Paper

Concept:In a geometric progression (GP), consecutive terms have a common ratio $r$, so $\frac{a_{p+1}}{a_p}=r$, and $\ln a_{p+1} - \ln a_p = \ln r$.When two columns of a determinant become identical, the determinant value is zero.Explanation:Let the GP have common ratio $r$. Represent the determinant as $D = \begin{vmatrix} \ln a_1 & \ln a_2 & \ln a_3 \\ \ln a_4 & \ln a_5 & \ln a_6 \\ \ln a_7 & \ln a_8 & \ln a_9 \end{vmatrix}$.Apply column operations: $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$.$D = \begin{vmatrix} \ln a_1 & \ln a_2 - \ln a_1 & \ln a_3 - \ln a_2 \\ \ln a_4 & \ln a_5 - \ln a_4 & \ln a_6 - \ln a_5 \\ \ln a_7 & \ln a_8 - \ln a_7 & \ln a_9 - \ln a_8 \end{vmatrix}$.Each difference simplifies: $\ln a_{p+1} - \ln a_p = \ln\left( \frac{a_{p+1}}{a_p} \right) = \ln r$.Thus $D = \begin{vmatrix} \ln a_1 & \ln r & \ln r \\ \ln a_4 & \ln r & \ln r \\ \ln a_7 & \ln r & \ln r \end{vmatrix}$.Now column $C_2$ and column $C_3$ are identical (both contain $\ln r$ in every row).A determinant with two identical columns equals zero.Hence the value of the given determinant is $0$.Answer:$0$ (Option A).