UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The product of the first, fourth, and fifth terms of an arithmetic progression is maximized for a specific common difference, from which the first term can be found.Explanation:In an arithmetic progression with first term $a$ and common difference $d$, the fourth term is $a + 3d$ and the fifth term is $a + 4d$.The product $P = a \cdot (a + 3d) \cdot (a + 4d)$ attains its maximum value when $d = \frac{8}{5}$ (derived from the condition of maximum product).From the given relation (obtained in the earlier step of the solution), we have $a = 2 - 5d$.Substitute $d = \frac{8}{5}$ into this relation:$a = 2 - 5 \times \frac{8}{5} = 2 - 8 = -6$.Thus, the first term is $-6$.Answer:$-6$ (Option B)