UPSC NDA II 2015 Mathematics Paper

Concept:In any triangle, the semi-perimeter is $s = \frac{a+b+c}{2}$ and the half-angle sine formulas are:$\sin\frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}$ and $\sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$.The condition that leads to sides being in arithmetic progression (A.P.) is $a+c = 2b$.Explanation:Start with the given equation:$2a \sin^2\frac{C}{2} + 2c \sin^2\frac{A}{2} = 2a + 2c - 3b$.Substitute the half-angle formulas:$2a \cdot \frac{(s-a)(s-b)}{ab} + 2c \cdot \frac{(s-b)(s-c)}{bc} = 2a+2c-3b$.Simplify each term: $\frac{2}{b}(s-a)(s-b) + \frac{2}{b}(s-b)(s-c) = 2a+2c-3b$.Factor $\frac{2}{b}(s-b)$: $\frac{2}{b}(s-b)[(s-a)+(s-c)] = 2a+2c-3b$.Note that $(s-a)+(s-c) = 2s - (a+c) = (a+b+c) - (a+c) = b$.So the left side becomes $\frac{2}{b}(s-b) \cdot b = 2(s-b)$.Thus $2(s-b) = 2a+2c-3b$.Replace $s$ with $\frac{a+b+c}{2}$: $2\left(\frac{a+b+c}{2} - b\right) = 2a+2c-3b$Simplify: $a+b+c - 2b = 2a+2c-3b$$a - b + c = 2a+2c-3b$Bring all terms to one side: $a - b + c - 2a - 2c + 3b = 0$Combine: $-a + 2b - c = 0$Thus $a + c = 2b$.This condition means the sides $a, b, c$ are in arithmetic progression (A.P.).Answer:B. A.P.