Question

The arithmetic mean of two numbers is $6$ and their geometric mean $G$ and harmonic mean $H$ satisfy the relation $G^2+3H=48$. Find the two numbers.

Answer 1

Let two numbers be $a,b$.

Arithmetic mean is $6$.

$\therefore \frac{a+b}{2}=6\Rightarrow a+b=\mathrm{12.....}\left(1\right)$

$G=(ab{)}^{1/2}$

$H=\frac{2ab}{a+b}$

Given: $G^2+3H=48\Rightarrow \left(\right(ab{)}^{1/2}{)}^2+\frac{3\times 2ab}{a+b}=48$

$ab+\frac{6ab}{a+b}=48$

Put value of $a+b$ by (1),

$ab+\frac{6ab}{12}=48\Rightarrow ab=32$

So, $a+b=12$ and $ab=32$ means $a=4$ and $b=8$.

Hence two numbers are $4$ and $8$.