UPSC NDA II 2023 Mathematics Solved Paper

Concept:If terms form a geometric progression (GP), each term is given by $t_n = a r^{n-1}$, where $a$ is the first term and $r$ is the common ratio.Explanation:The given terms are $t_1, t_3, t_5, \ldots, t_{21}$.These are every alternate term starting from $t_1$, so there are $11$ terms in total.Let $P = (t_1 t_3 t_5 \cdots t_{21})^{1/11}$.Writing each term using $a$ and $r$:$t_1 = a$, $t_3 = a r^2$, $t_5 = a r^4$, $\ldots$, $t_{21} = a r^{20}$.Thus, $P = (a \cdot a r^2 \cdot a r^4 \cdots a r^{20})^{1/11}$.There are $11$ factors of $a$, so the product becomes $a^{11} \cdot r^{2+4+6+\cdots+20}$.The exponent sum is $2(1+2+3+\cdots+10) = 2 \times \frac{10 \times 11}{2} = 110$.Hence, $P = (a^{11} r^{110})^{1/11} = a r^{10}$.Since $t_{11} = a r^{10}$, we obtain $P = t_{11}$.Answer:$t_{11}$ (Option C).