UPSC NDA 2 2022 Mathematics Solved Paper

Concept:To find the ratio of the first terms of two arithmetic progressions, set the term index $n = 1$ in the given sum ratio, which isolates the first terms.Explanation:Let the first term of A be $a$ and common difference $d$.Let the first term of B be $M$ and common difference $D$.Sum of $n$ terms of A: $P = \frac{n}{2}[2a + (n-1)d]$.Sum of $n$ terms of B: $Q = \frac{n}{2}[2M + (n-1)D]$.Given: $\frac{P}{Q} = \frac{5n+4}{9n+6}$.Cancel $\frac{n}{2}$: $\frac{2a+(n-1)d}{2M+(n-1)D} = \frac{5n+4}{9n+6}$.Divide numerator and denominator by $2$: $\frac{a+\frac{n-1}{2}d}{M+\frac{n-1}{2}D} = \frac{5n+4}{9n+6}$.To obtain the ratio $\frac{a}{M}$, set $\frac{n-1}{2} = 0$, i.e., $n = 1$.Substitute $n=1$: $\frac{a}{M} = \frac{5(1)+4}{9(1)+6} = \frac{9}{15} = \frac{3}{5}$.Thus, the ratio of the first term of A to that of B is $\frac{3}{5}$.Answer:$\frac{3}{5}$