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Question

The geometric mean of two numbers exceeds by $$12$$ the smaller of the numbers, and the arithmetic mean of the same numbers is smaller by $$24$$ than the larger of the numbers. Find the numbers.


Solution

Let the two numbers be $$a$$ and $$b$$
$$GM=\sqrt { ab } $$ and $$AM=\cfrac { a+b }{ 2 } $$
$$\sqrt { ab } =12+a\quad -(i)\\ \cfrac { a+b }{ 2 } =b-24\quad -(ii)\\ a+b=2b-48\\ b=a+48$$
Putting value of $$b$$ in $$(i)$$
$$ \sqrt { a(a+48) } =12+a$$
Squaring both sides
$$ { a }^{ 2 }+48a={ a }^{ 2 }+24a+144\\ \Rightarrow 24a=144\\ a=\cfrac { 144 }{ 24 } =6\\ So,b=48+6\\ =54$$
Numbers are $$54\quad \&\quad 6$$

Mathematics

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