Mathematics - Sequence & Series Question with Solution | TestHub

Let the unknown number be $b$ and the H.M. be $H$. Then $H=\frac{2\cdot \frac{1}{3}\cdot b}{\frac{1}{3}+b}=\frac{2b}{1+3b}\dots (1)$ $\frac{1}{H}-6=\sqrt{3\cdot \frac{1}{b}}\dots (2)$. From (1): $\frac{1+3b}{2b}-6=\frac{1-9b}{2b}$, so $\frac{1-9b}{2b}=\sqrt{\frac{3}{b}}$. Squaring: $\frac{(1-9b{)}^2}{4b^2}=\frac{3}{b}\Rightarrow 81b^2-18b+1=12b\Rightarrow 81b^2-30b+1=0\Rightarrow (27b-1)(3b-1)=0$. $b=\frac{1}{27}$ or $\frac{1}{3};\frac{1}{3}$ is rejected because the terms are distinct. So $b=\frac{1}{27}$ and $H=\frac{2/27}{1+1/9}=\frac{1}{15}$.