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)

$9\left(25a^2+b^2\right)+25\left(c^2+3ac\right)=15b\left(3a+c\right)$
$\Rightarrow {\left(15a\right)}^2+{\left(3b\right)}^2+{\left(5c\right)}^2-45ab-15bc-75ac=0$
$\Rightarrow {\left(15a-3b\right)}^2+{\left(3b-5c\right)}^2+{\left(15a-5c\right)}^2=0$
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.