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  4. Find two positive numbers whose difference is 12 and whose A.M. exceeds their G.M. by 2.

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Q. Find two positive numbers whose difference is $12$ and whose $A.M$. exceeds their $G.M$. by $2$.

Sequences and Series

Solution:

Let the two numbers be $a$ and $b$. Then,
$ a-b =12\quad...\left(i\right)$
It is given that $A.M. - G.M. = 2 $
$\Rightarrow \frac{a+b}{2} - \sqrt{ab} = 2$
$ \Rightarrow a+b -2\sqrt{ab} =4$
$\Rightarrow \left(\sqrt{a}-\sqrt{b}\right)^{2} = 4 $
$ \Rightarrow \sqrt{a}-\sqrt{b}= \pm2\quad...\left(ii\right)$
Now, $a-b = 12 $
$ \Rightarrow \left(\sqrt{a} +\sqrt{b}\right) \left(\sqrt{a} -\sqrt{b}\right) = 12$
$\Rightarrow \left(\sqrt{a } +\sqrt{b}\right)\times \left(\pm2\right) = 12 \quad$ [Using $(ii)$]
$ \Rightarrow \sqrt{a} + \sqrt{b} = \pm 6 \quad ...\left(iii\right)$
Solving $\left(ii\right)$ and $\left(iii\right)$, we get $a= 16, b=4$