  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 $$3$$.Maths

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