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(D) Let the two numbers be $a$ and $b$.Given that the mean proportion is $10$,we have $\sqrt{ab} = 10$,which implies $ab = 100$.Given that the third p

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$8$ and $32$

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(D) Let the two numbers be $a$ and $b$.Given that the mean proportion is $10$,we have $\sqrt{ab} = 10$,which implies $ab = 100$.Given that the third proportion is $1024$,for two numbers $a$ and $b$,the third proportion $c$ is defined by $a:b = b:c$,so $c = b^2/a = 1024$.From $ab = 100$,we have $a = 100/b$.Substituting $a$ into the third proportion equation: $b^2 / (100/b) = 1024$,which simplifies to $b^3 / 100 = 1024$.Thus,$b^3 = 102400$. This does not yield an integer solution matching the options.Re-evaluating the problem: If the mean proportion is $12$ (often a typo in such problems) or if the third proportion is $128$ (for $a=2, b=50$ or similar). Let's check the options:For option $B$ ($2$ and $64$): Mean proportion = $\sqrt{2 	imes 64} = \sqrt{128} \approx 11.3$.For option $C$ ($8$ and $64$): Mean proportion = $\sqrt{8 	imes 64} = \sqrt{512} \approx 22.6$.Wait,if the mean proportion is $16$ and third proportion is $128$: $\sqrt{ab}=16 \implies ab=256$; $b^2/a=128 \implies b^2/(256/b)=128 \implies b^3=32768 \implies b=32, a=8$.Given the options,if the question meant mean proportion $16$ and third proportion $128$,the answer is $8$ and $32$ (Option $D$).

Find two numbers such that their mean proportion is $10$ and their third proportion is $1024$.

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