If A.M. and G.M. of two positive numbers a an | vedclass

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(A) Given that $A.M. = \frac{a+b}{2} = 10 \implies a+b = 20$ $(1)$Given that $G.M. = \sqrt{ab} = 8 \implies ab = 64$ $(2)$Using the identity $(a-b)^2

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$4, 16$

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(A) Given that $A.M. = \frac{a+b}{2} = 10 \implies a+b = 20$ $(1)$Given that $G.M. = \sqrt{ab} = 8 \implies ab = 64$ $(2)$Using the identity $(a-b)^2 = (a+b)^2 - 4ab$,we get:$(a-b)^2 = (20)^2 - 4(64) = 400 - 256 = 144$Thus,$a-b = \pm 12$ $(3)$Solving $(1)$ and $(3)$:If $a-b = 12$,then $2a = 32 \implies a = 16$ and $b = 4$.If $a-b = -12$,then $2a = 8 \implies a = 4$ and $b = 16$.Therefore,the numbers are $4$ and $16$.

If $A.M.$ and $G.M.$ of two positive numbers $a$ and $b$ are $10$ and $8$ respectively,find the numbers.

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