Mathematics - Sequence & Series Question with Solution | TestHub

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

Given that $\frac{a+b}{2}=2\sqrt{ab}$. Squaring both sides, we get $(a+b{)}^2=16ab$.

Thus, $a^2+2ab+b^2=16ab$, which simplifies to $a^2-14ab+b^2=0$. Dividing by $b^2$, we have $(\frac{a}{b}{)}^2-14(\frac{a}{b})+1=0$.

Solving for $\frac{a}{b}$ using the quadratic formula: $\frac{a}{b}=\frac{14\pm \sqrt{196-4}}{2}=7\pm \sqrt{48}=7\pm 4\sqrt{3}=(2\pm \sqrt{3}{)}^2$.

Therefore, $\frac{a}{b}=\frac{(2+\sqrt{3}{)}^2}{1}$ or $\frac{(2-\sqrt{3}{)}^2}{1}$.

Taking the square root, $\sqrt{\frac{a}{b}}=2+\sqrt{3}$ or $2-\sqrt{3}$. Thus, $a:b=2+\sqrt{3}:2-\sqrt{3}$.

Final Answer: The final answer is $2+\sqrt{3}:2-{\sqrt{3}}_{}$