Question

The geometric mean of two numbers exceeds by $12$ the smaller of the numbers, and the arithmetic mean of the same numbers is smaller by $24$ than the larger of the numbers. Find the numbers.

Answer 1

Let the two numbers be $a$ and $b$

$GM=\sqrt{ab}$ and $AM=\frac{a+b}{2}$

$\sqrt{ab}=12+a-\left(i\right)\frac{a+b}{2}=b-24-\left(ii\right)a+b=2b-48b=a+48$

Putting value of $b$ in $\left(i\right)$

$\sqrt{a(a+48)}=12+a$

Squaring both sides

$a^2+48a=a^2+24a+144\Rightarrow 24a=144a=\frac{144}{24}=6So,b=48+6=54$

Numbers are $54\&6$