Question

If $a,b$ and $c$ are positive real number such that $9(25a^2+b^2)+25(c^2–3ac)=15b(3a+c)$, then which of the following is INCORRECT?

• A. $a,b$ and $c$ are in A.P.
• B. $b,c$ and $a$ are in A.P.
• C. $a,2c-a$ and $2b-a$ are in A.P.
• D. $a+1,c+1$ and $b+1$ are in A.P.

Answer 1

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

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