Question

The H.M. of two number is $4$ and their $A$ $m$. and $G.M.$ satisfy the relation $2A+G^2=27$, then the numbers are :

Answer 1

Let the two numbers be a and b, then

$\frac{2ab}{a}+b=4\dots .\left(1\right);a+\frac{b}{2}=A;\sqrt{ab}=G$

Also $2A+G^2=27\Rightarrow a+b+ab=27$ ………… (2)

Putting $ab=27–(a+b)$ in eqn. (1). We get

$54–2\frac{(a+b)}{a}+b=4\Rightarrow a+b=9$ then $ab=27–9=18$

Solving the two we gat $a=6,b=3$ or $a=3,b=6$, which are the required numbers.