UPSC NDA II 2025 Mathematics Paper

Concept:The common terms of two arithmetic progressions (APs) satisfy both general formulas. We find the condition for equality and count valid terms within the given limits.Explanation:For AP1: first term 19, common difference 2, 110 terms. Its $n^{\text{th}}$ term is $T_n = 19 + 2(n-1) = 2n + 17$, with last term $T_{110} = 237$. For AP2: first term 19, common difference 3, 75 terms. Its $m^{\text{th}}$ term is $S_m = 19 + 3(m-1) = 3m + 16$, with last term $S_{75} = 241$. Set the terms equal: $2n + 17 = 3m + 16 \Rightarrow 2n - 3m = -1$. Solve modulo 3: $2n \equiv 2 \ (\text{mod } 3) \Rightarrow n \equiv 1 \ (\text{mod } 3)$. So $n = 3k + 1$ for integer $k$. The common term is $T_{3k+1} = 2(3k+1) + 17 = 6k + 19$. This term must not exceed the last term of AP1 (237): $6k + 19 \le 237 \Rightarrow 6k \le 218 \Rightarrow k \le 36.33$, so $k \le 36$. Also $k \ge 0$. Thus $k$ takes values 0, 1, 2, ..., 36, giving 37 common terms. Answer:37 (Option C)