Question

The harmonic mean of two numbers is $4$, their A.M. $A$, and G.M. $G$. satisfy the relation $2A+G^2=27.$ Find a? (Let the two numbers be a,b where a<b)

Answer 1

Let required number are $a$ and $b$
So using given conditions,
$A=\frac{a+b}{2}ora+b=2A,$
$G=\sqrt{ab}\therefore G^2=ab$
or $H=\frac{2ab}{a+b}=4,\therefore G^2=AH$ gives $G^2=4A.$
Also $2A+G^2=27or2A+4A=27$
$\therefore A=\frac{27}{6}=\frac{9}{2}$
$\therefore \frac{a+b}{2}=\frac{9}{2}ora+b=9$ ...(1)
Also $G^2=4A=4.\frac{9}{2}=18orab=18$ ...(2)
From (1) and (2) we conclude that a and b are the roots of
$t^2-9t+18=0or(t-6)(t-3)=0$
$\therefore t=6,3$
Hence the numbers are $6$ and $3$.