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

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(C) Given equation: $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$Expanding the terms: $225a^2 + 9b^2 + 25c^2 - 75ac = 45ab + 15bc$Rearranging the ter

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

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(C) Given equation: $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$Expanding the terms: $225a^2 + 9b^2 + 25c^2 - 75ac = 45ab + 15bc$Rearranging the terms: $225a^2 + 9b^2 + 25c^2 - 45ab - 15bc - 75ac = 0$Multiplying by $2$: $450a^2 + 18b^2 + 50c^2 - 90ab - 30bc - 150ac = 0$This can be written as: $(15a - 3b)^2 + (3b - 5c)^2 + (5c - 15a)^2 = 0$For the sum of squares to be zero,each term must be zero:$15a - 3b = 0 \Rightarrow 3b = 15a \Rightarrow b = 5a$$3b - 5c = 0 \Rightarrow 3b = 5c$$5c - 15a = 0 \Rightarrow 5c = 15a \Rightarrow c = 3a$Now,check the condition for $A.P.$ for $b, c, a$:$2c = b + a \Rightarrow 2(3a) = 5a + a \Rightarrow 6a = 6a$Since $2c = a + b$,the numbers $b, c, a$ are in $A.P.$

For any three positive real numbers $a, b, c$,if $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$,then:

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