Question

If A, G & 4 are A.M., G.M & H.M. of two numbers respectively and $2A+G^2=27$, then the numbers are

• A. 8, 2
• B. 3, 6
• C. 6, 3
• D. 6, 4

Answer 1

Answer: (B)

3, 6

we have,

$HM=\frac{2ab}{a+b}=4$

$\frac{2G^2}{2A}=4$

$\Rightarrow G^2=4A$

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

$\Rightarrow 2A=a+b$

$GM=\sqrt{ab}=G$

$\Rightarrow G^2=ab$

Given,

$2A+G^2=27$

$2A+4A=27$

$6A=27$

$6\left(\frac{a+b}{2}\right)=27$

$\Rightarrow a+b=9$..........(1)

$\frac{2ab}{a+b}=4\Rightarrow \frac{2ab}{9}=4$

$ab=18$

$b=\frac{18}{a}$...............(2)

substituting (2) in (1), we get,

$a+\frac{18}{a}=9$

$\Rightarrow a^2-9a+18=0$

$(a-3)(a-6)=0$

$a=3,6$

when $a=3,b=\frac{18}{3}=6$

when $a=6,b=\frac{18}{6}=3$

$\therefore 3,6$