UPSC NDA 2 2022 Mathematics Solved Paper

Concept:We find the ratio of the 10th terms of two arithmetic progressions using the given ratio of their sums of $n$ terms.Explanation:Let the first AP have first term $a$ and common difference $d$.Its $n$th term is $a + (n-1)d$ and sum of $n$ terms is $\frac{n}{2}[2a + (n-1)d]$.Let the second AP have first term $M$ and common difference $D$.Its $n$th term is $M + (n-1)D$ and sum of $n$ terms is $\frac{n}{2}[2M + (n-1)D]$.Given: $\frac{\text{Sum of }n\text{ terms of first AP}}{\text{Sum of }n\text{ terms of second AP}} = \frac{5n+4}{9n+6}$.Cancel $\frac{n}{2}$ from numerator and denominator:$\frac{2a + (n-1)d}{2M + (n-1)D} = \frac{5n+4}{9n+6}$.Divide numerator and denominator of left side by 2:$\frac{a + \frac{n-1}{2}d}{M + \frac{n-1}{2}D} = \frac{5n+4}{9n+6}$ ... (1)We need ratio of 10th terms: $\frac{a+9d}{M+9D}$.Notice that the coefficient of $d$ in numerator of (1) is $\frac{n-1}{2}$, while we need coefficient 9.Set $\frac{n-1}{2}=9 \Rightarrow n=19$.Substitute $n=19$ in (1):$\frac{a+9d}{M+9D} = \frac{5(19)+4}{9(19)+6} = \frac{95+4}{171+6} = \frac{99}{177}$.Simplify: divide numerator and denominator by 3: $\frac{33}{59}$.Thus the required ratio is $\frac{33}{59}$.The correct option is C.Answer:$\frac{33}{59}$ (Option C)