UPSC NDA 2 2016 Mathematics Solved Paper

Concept:In an arithmetic progression (AP), the middle term is the average of the first and last terms. In a harmonic progression (HP), the reciprocals form an AP. For three terms in AP: $2 \times \text{(middle)} = \text{first} + \text{last}$.Explanation:Given: $a, x, y, z, b$ are in AP. Since they are equally spaced, $y$ is the middle term of this 5‑term AP. Thus $a + b = 2y$. Also, the sum $x + y + z = 15$. In a 3‑term subsequence $(x, y, z)$, $2y = x + z$. So $x+y+z = (2y) + y = 3y = 15$, giving $y = 5$. Hence $a + b = 2 \times 5 = 10$. …(1) Now, $a, p, q, r, b$ are in HP. Reciprocals $\frac1a, \frac1p, \frac1q, \frac1r, \frac1b$ are in AP. Here $\frac1q$ is the middle term of the 5‑term AP, so $\frac1a + \frac1b = 2 \cdot \frac1q$. Also, for the three reciprocals $\frac1p, \frac1q, \frac1r$, we have $2 \cdot \frac1q = \frac1p + \frac1r$. Given: $\frac1p + \frac1q + \frac1r = \frac53$. Substituting $\frac1p + \frac1r = 2 \cdot \frac1q$, we get $2 \cdot \frac1q + \frac1q = \frac53$, so $3 \cdot \frac1q = \frac53$, hence $\frac1q = \frac59$. Thus $\frac1a + \frac1b = 2 \cdot \frac59 = \frac{10}{9}$. Write $\frac{a+b}{ab} = \frac{10}{9}$. Using (1) $a+b = 10$, we get $\frac{10}{ab} = \frac{10}{9}$. Therefore $ab = 9$.Answer:$ab = 9$ (Option B).