Question

The arithmetic mean of two positive numbers $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean exceeds their harmonic mean by $\frac{6}{5}.$ If $a+b=\alpha$ and $\left|a-b\right|=\beta ,$ then the value of $\frac{10\beta }{\alpha }$ is equal to

Answer 1

Answer: 6

$A-G=\frac{3}{2}$ and $G-H=\frac{6}{5}$
Also, we know that $G^2=AH$
$\Rightarrow G^2=\left(\frac{3}{2}+G\right)\left(G-\frac{6}{5}\right)$
$\Rightarrow \left(\frac{3}{2}-\frac{6}{5}\right)G=\frac{9}{5}\Rightarrow G=\frac{9}{5}\times \frac{10}{3}=6$
$G=6\Rightarrow A=\frac{15}{2}$
$\Rightarrow \frac{a+b}{2}=\frac{15}{2}$ and $\sqrt{ab}=6$
$\Rightarrow a+b=15$ and $ab=36$
$\left|a-b\right|=\sqrt{{\left(a+b\right)}^2-4ab}=\sqrt{225-4\times 36}=9$
$\alpha =a+b=15,\beta =\left|a-b\right|=9$
$\Rightarrow \frac{10\beta }{\alpha }=\frac{10\times 9}{15}=6$