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Question

If the $$A.M.$$ between two numbers exceeds their $$G.M.$$ by $$2$$ and the $$G.M.$$ exceeds their $$H.M.$$ by $$8/5$$, find the smaller of the two numbers.


Solution

If the numbers be a and b, then
$$\displaystyle A-G= 2 \therefore \left ( a+b \right )/2-\sqrt{ab}= 2$$
or $$\displaystyle \left ( \sqrt{a}-\sqrt{b} \right )^{2}= 4$$ ...(1)
$$\displaystyle G-H= \frac{8}{5} \therefore \sqrt{ab}-\frac{2ab}{a+b}= \frac{8}{5}$$
or $$\displaystyle \sqrt{ab}\frac{\left [ a+b-2\sqrt{ab} \right ]}{a+b}= \frac{8}{5}$$
or $$\displaystyle 5\sqrt{ab}\left ( \sqrt{a}-\sqrt{b} \right )^{2}= 8\left ( a+b \right )$$
or $$\displaystyle 5\sqrt{ab}.4= 8\left ( a+b \right ),$$ by (1)
$$\displaystyle \therefore 2a-5\sqrt{ab}+2b+0$$
or $$\displaystyle \left ( 2\sqrt{a}-\sqrt{b} \right )\left ( \sqrt{a}-2\sqrt{b} \right )= 0$$
$$\displaystyle \therefore \sqrt{a}= 2\sqrt{b}\:or\:\left ( \sqrt{b}/2 \right ) \therefore a= 4b\:or\:\left ( b/4 \right )$$
Form (1), $$\displaystyle \left ( 2\sqrt{b}-\sqrt{b} \right )^{2}= 4\:or\:b= 4$$
$$\displaystyle \therefore \sqrt{a}= 2\sqrt{b}= 4\:or\:a= 16$$
Hence the numbers are $$16$$ and $$4$$. The other factor will give numbers as $$4$$ and $$16$$.

Mathematics

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