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Question

If $$a,b,c$$ are three distinct positive real numbers which are in H.P then $$\cfrac { 3a+2b }{ 2a-b } +\cfrac { 3c+2b }{ 2c-b } $$ is


A
Greater than or equal to 10
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B
Less than or equal to 10
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C
Only equal to 20
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D
None of these
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Solution

The correct option is A Greater than or equal to $$10$$
$$a,b,c$$ in H.P, $$b=\dfrac{2ac}{a+c}...(1)$$
$$\dfrac{3a+2b}{2a-b}+\dfrac{3c+2b}{2c-b}=\dfrac{6ac+4bc-3ab-2b^2+6ac+4ab-3bc-2b^2}{4ac-2ab-2bc+b^2}$$
$$=\dfrac{12ac+b(a+c)-4b^2}{4ac-2b(a+c)+b^2}$$
Replace $$b(a+c)=2ac$$ by relation (1)
$$=\dfrac{12ac+2ac-4b^2}{4ac-4ac+b^2}=\dfrac{14ac-4b^2}{b^2}$$
$$=\dfrac{14ac}{b^2}-4$$
Replace $$\dfrac{2ac}{b}=a+c$$,
$$\dfrac{7(a+c)}{b}-4$$
As $$a,b,c$$ in H.P $$\therefore \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$$
$$\Rightarrow \dfrac{7(a+c)}{b}-4=7(a+c)\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)-4$$
$$=\dfrac{7}{2}\left(1+\dfrac{a}{c}+\dfrac{c}{a}+1\right)-4$$
$$=\dfrac{7}{2}\left(2+\dfrac{a}{c}+\dfrac{c}{a}\right)-4=3+\dfrac{7}{2}\left(\dfrac{a}{c}+\dfrac{c}{a}\right)$$
By Arithmetic mean relation on $$\dfrac{a}{c},\dfrac{c}{a}$$
$$\Rightarrow \dfrac{\left(\dfrac{a}{c}+\dfrac{c}{a}\right)}{2}\geq \sqrt{\dfrac{a}{c}\times\dfrac{c}{a}} $$
$$\dfrac{a}{c}+\dfrac{c}{a}\geq 2$$
$$\therefore 3+\dfrac{7}{2}\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\geq 3+\dfrac{7}{2}\times 2=10$$
Will be greater than or equal to 10.


Mathematics

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