Question

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

• A. $b, c$ and $a$ are in $A.P$
• B. $a, b$ and $c$ are in $A.P$
• C. $a, b$ and $c$ are in $G.P$
• D. $b, c$ and $a$ are in $G.P$

Answer 1

Answer: (A)

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

Answer 2

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