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

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(C) Let the two numbers be $a$ and $b$.Given that the ratio of the numbers is $4:1$,we have $a = 4b$.We know that $A.M. = \frac{a+b}{2}$ and $G.M. = \

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$16, 4$

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(C) Let the two numbers be $a$ and $b$.Given that the ratio of the numbers is $4:1$,we have $a = 4b$.We know that $A.M. = \frac{a+b}{2}$ and $G.M. = \sqrt{ab}$.According to the problem,$A.M. = G.M. + 2$.Substituting the values,we get $\frac{4b+b}{2} = \sqrt{4b 	imes b} + 2$.$\frac{5b}{2} = \sqrt{4b^2} + 2$.$\frac{5b}{2} = 2b + 2$.Multiplying by $2$,we get $5b = 4b + 4$.$b = 4$.Since $a = 4b$,$a = 4(4) = 16$.Thus,the numbers are $16$ and $4$.

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

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