If the A.M. of two numbers is greater than G. | vedclass

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Question

(C) Let the two numbers be $a$ and $b$.The Arithmetic Mean $(A.M.)$ is given by $\frac{a + b}{2}$.The Geometric Mean $(G.M.)$ is given by $\sqrt{ab}$.

Vedclass · Accepted Answer

$16, 4$

Vedclass · Answer

(C) Let the two numbers be $a$ and $b$.The Arithmetic Mean $(A.M.)$ is given by $\frac{a + b}{2}$.The Geometric Mean $(G.M.)$ is given by $\sqrt{ab}$.According to the problem,$A.M. = G.M. + 2$,so $\frac{a + b}{2} = \sqrt{ab} + 2$ ......$(i)$.The ratio of the numbers is $4:1$,so $\frac{a}{b} = \frac{4}{1}$,which implies $a = 4b$ ......$(ii)$.Substitute $a = 4b$ into equation $(i)$:$\frac{4b + b}{2} = \sqrt{4b \cdot b} + 2$$\frac{5b}{2} = \sqrt{4b^2} + 2$$\frac{5b}{2} = 2b + 2$Multiply by $2$: $5b = 4b + 4$,which gives $b = 4$.Now,substitute $b = 4$ into equation $(ii)$:$a = 4(4) = 16$.Thus,the numbers are $16$ and $4$.

If the $A.M.$ of two numbers is greater than $G.M.$ of the numbers by $2$ and the ratio of the numbers is $4:1$,then the numbers are

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