UPSC NDA 2 2022 Mathematics Solved Paper

Concept:The ratio of sums of $n$ terms of two APs is given. Using this ratio, we find the ratio of first terms and the ratio of second terms. Then we compare the common differences $d$ and $D$.Explanation:Let the first AP have first term $a$ and common difference $d$. Let the second AP have first term $M$ and common difference $D$. Given: $\frac{\text{Sum of } n \text{ terms of first AP}}{\text{Sum of } n \text{ terms of second AP}} = \frac{5n+4}{9n+6}$. Using sum formula: $\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D]} = \frac{5n+4}{9n+6}$. Cancel $\frac{n}{2}$: $\frac{2a+(n-1)d}{2M+(n-1)D} = \frac{5n+4}{9n+6}$. Rewrite numerator as $a + \frac{n-1}{2}d$ and denominator as $M + \frac{n-1}{2}D$. So $\frac{a + \frac{n-1}{2}d}{M + \frac{n-1}{2}D} = \frac{5n+4}{9n+6}$ ... (1). To get the ratio of first terms, set $\frac{n-1}{2}=0$ i.e. $n=1$. From (1): $\frac{a}{M} = \frac{5(1)+4}{9(1)+6} = \frac{9}{15} = \frac{3}{5}$. So $a = 3k$, $M = 5k$ for some $k>0$. To get the ratio of second terms, set $\frac{n-1}{2}=1$ i.e. $n=3$. From (1): $\frac{a+d}{M+D} = \frac{5(3)+4}{9(3)+6} = \frac{19}{33}$. Substitute $a=3k$, $M=5k$: $\frac{3k+d}{5k+D} = \frac{19}{33}$. Cross multiply: $33(3k+d) = 19(5k+D)$. $99k + 33d = 95k + 19D$. $4k + 33d = 19D$. Thus $D = \frac{4k+33d}{19}$. Now compare $D$ and $d$: $D - d = \frac{4k+33d}{19} - d = \frac{4k+33d-19d}{19} = \frac{4k+14d}{19} = \frac{2(2k+7d)}{19}$. Since $k>0$ and $d>0$ (common difference of an AP is positive here), $2k+7d > 0$, so $D - d > 0$. Hence $D > d$ always.Answer:$D > d$ is always correct. Thus option (A) is the correct choice.