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

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