For any three positive real numbers a,b,c ; 9 | vedclass

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Question

We have 

$9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$

$ \Rightarrow 225{a^2} + 9{b^2}

Vedclass · Accepted Answer

$b,c,a$ are in $A.P.$

Vedclass · Answer

We have 

$9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$

$ \Rightarrow 225{a^2} + 9{b^2} + 25{c^2} - 75ac = 45ab + 15bc$

$ \Rightarrow {\left( {15a} \right)^2} + {\left( {3b} \right)^2} + 5{c^2} - 75ac - 45ab - 15bc = 0$

$\frac{1}{2}\left[ {{{\left( {15a - 3b} \right)}^2} + {{\left( {3b - 5c} \right)}^2} + {{\left( {5c - 15a} \right)}^2}} \right] = 0$

it is possible when $15a - 3b = 0,3b - 5c = 0$ and $5c - 15a = 0$

$ \Rightarrow 15a = 3b = 5c$

$ \Rightarrow b\frac{{5c}}{3},a = \frac{c}{3}$

$ \Rightarrow a + b = \frac{c}{3} + \frac{{5c}}{3} = \frac{{6c}}{3}$

$ \Rightarrow a + b = 2c$

$ \Rightarrow b,c,a$ are in $A.P$

For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then

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