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Question

The harmonic mean of two numbers is $$4$$, their A.M. $$A$$, and G.M. $$G$$. satisfy the relation $$\displaystyle 2A+G^{2}= 27.$$ Find a? (Let the two numbers be a,b where a<b)


Solution

Let required number are $$ a$$ and $$b$$
So using given conditions,
$$\displaystyle A= \frac{a+b}{2}\:or\:a+b= 2A,$$
$$\displaystyle G= \sqrt{ab} \therefore G^{2}= ab$$
or $$\displaystyle H= \frac{2ab}{a+b}= 4,  \therefore G^{2}= AH$$ gives $$G^{2}= 4A.$$
Also $$\displaystyle 2A+G^{2}= 27\:or\:2A+4A= 27$$
$$\displaystyle \therefore A= \frac{27}{6}= \frac{9}{2}$$
$$\displaystyle \therefore \frac{a+b}{2}= \frac{9}{2}\:or\:a+b= 9$$ ...(1)
Also $$\displaystyle G^{2}= 4A= 4.\frac{9}{2}= 18\:or\:ab= 18$$ ...(2)
From (1) and (2) we conclude that a and b are the roots of
$$\displaystyle t^{2}-9t+18= 0\:or\:\left ( t-6 \right )\left ( t-3 \right )= 0$$
$$\displaystyle \therefore t= 6,3$$
Hence the numbers are $$6$$ and $$3$$.

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