UPSC NDA 2 2021 Mathematics Solved Paper

Concept:If $x, y, z$ are in GP, then $y^{2}=xz$. If three numbers are in AP, then the middle term equals the average of the other two. If three numbers are in HP, then their reciprocals are in AP.Explanation:Given: $x, y, z$ are in GP. So $y^{2}=xz$. For statement 1: Multiply both sides of $y^{2}=xz$ by $9$: $(3y)^{2}=9xz = (3x)(3z)$. Take natural logarithm: $\ln(3y)^{2} = \ln[(3x)(3z)]$. This gives $2\ln(3y) = \ln(3x) + \ln(3z)$. Therefore $\ln(3x), \ln(3y), \ln(3z)$ satisfy the AP condition. Thus statement 1 is correct. For statement 2: From $y^{2}=xz$, take $\ln$: $2\ln y = \ln x + \ln z$. Add $2xyz$ to both sides: $2(xyz+\ln y) = (xyz+\ln x)+(xyz+\ln z)$. So $xyz+\ln x,\ xyz+\ln y,\ xyz+\ln z$ are in AP. If three numbers are in AP, their reciprocals are in HP, not the numbers themselves. Therefore these three terms are not in HP. Statement 2 is false.Answer:Only statement 1 is correct. The correct option is A.