- Home
- Standard 11
- Mathematics
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
$a,b,c$ are in $G.P.$
$b,c,a$ are in $G.P.$
$b,c,a$ are in $A.P.$
$a,b,c$ are in $A.P.$
Solution
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 $A.P$
