Question

The harmonic mean of two numbers is 4 their A . M. A and G.M. G satisfies the relation $2A+G^2=27$. Find the two numbers

• A. $3$ and $3$
• B. $6$ and $3$
• C. $2$ and $3$
• D. $6$ and $6$

Answer 1

Answer: (B)

$6$ and $3$

$A=\frac{a+b}{2}ora+b=2A$

$G=\sqrt{a}b\therefore G^2=aborH=\frac{2ab}{a+b}=4\therefore G^2=AHGives{G}^2+4A$ Also ,

$2A+C^2=27Or2A+4A=27\therefore A=\frac{27}{6}=\frac{9}{6}\therefore \frac{a+b}{2}\frac{9}{2}ora+b=9$

Also $G^2=4A=4\dot{}\frac{9}{2}=ab=18$

From (1) and (2) we conclude that a and b are the root of

$t^2-9t+18=0or(t-6)(t-3)=0\therefore t=6,3$

Hence the number are $6$ and $3$