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

Answer 1

The correct option is B $b,c$ and $a$ are in A.P.

$9(25a^2+b^2)+25(c^2–3ac)=15b(3a+c)\Rightarrow 225a^2+9b^2+25c^2–75ac–45ab–15bc=0\Rightarrow (15a–3b{)}^2+(15a–5c{)}^2+(3b–5c{)}^2=0\Rightarrow 15a=3b=5c$
$\frac{a}{1}=\frac{b}{5}=\frac{c}{3}=\lambda \Rightarrow a=\lambda ,b=5\lambda ,c=3\lambda 2c=a+b$
So $b,c\text{and}a$ are in A.P.