Question

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

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

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