Question

The harmonic mean of two positive numbers $a$ and $b$ is $4,$ their arithmeitc mean is $A$ and the geometric mean is $G.$ If $2A+G^2=27,$ $a+b=\alpha$ and $\left|a-b\right|=\beta$ , then the value of $\frac{\alpha }{\beta }$ is equal to

• A. $1$
• B. $3$
• C. $\frac{5}{2}$
• D. $5$

Answer 1

Answer: (B)

Given, harmonic mean $H=4$
We know that, $G^2=AH$
Since, $2A+G^2=27$
$\Rightarrow 2A+AH=27$
$\Rightarrow 2A+4A=27$
$\Rightarrow A=\frac{27}{6}=\frac{9}{2}=\frac{a+b}{2}\Rightarrow a+b=9$
$G^2=AH=\frac{9}{2}\times 4=18$
$\Rightarrow ab=18$
$\left|a-b\right|=\sqrt{{\left(a+b\right)}^2-4ab}=\sqrt{81-4\times 18}=3$
$\Rightarrow \alpha =9,\beta =3$
$\Rightarrow \frac{\alpha }{\beta }=3$