UPSC NDA I 2022 Mathematics Solved Paper

Concept:If $a$, $b$, $c$ are in geometric progression (GP) with $a,b,c > 0$, then $b^2 = ac$. This property extends naturally to squares, reciprocals, and square roots.Explanation:Given $a, b, c > 0$ are in GP, so $b^2 = ac$.For statement 1: $a^2, b^2, c^2$ are in GP because $(b^2)^2 = b^4 = (ac)^2 = a^2 c^2$, so the geometric mean condition holds.For statement 2: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in GP. Let the common ratio be $r$, so $b = ar$ and $c = ar^2$. Then $\frac{1/b}{1/a} = \frac{a}{b} = \frac{1}{r}$ and $\frac{1/c}{1/b} = \frac{b}{c} = \frac{1}{r}$, giving a constant common ratio.For statement 3: $\sqrt{a}, \sqrt{b}, \sqrt{c}$ are in GP. From $b^2 = ac$, taking square roots gives $b = \sqrt{ac}$, so $\sqrt{b} = \sqrt{\sqrt{a}\sqrt{c}}$, which means $\sqrt{b}$ is the geometric mean of $\sqrt{a}$ and $\sqrt{c}$.Thus all three statements are correct.Answer:Statements 1, 2 and 3 are correct.