Question

Consider two positive numbers $a$ and $b$. If arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$, then the absolute value of $\left(a^{2}-b^{2}\right)$ is equal to

Answer 1

$A=G+\frac{3}{2}, G=H+\frac{6}{5}$
Using $G^{2}=A H$ we get
$G=6, A=\frac{15}{2}$ and $H=\frac{24}{5}$
Hence, $a+b=15$ and $a b=36$
$a=12, b=3$ or $a=3, b=12$
$\left|a^{2}-b^{2}\right|=135$.