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

The correct option is A

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

Explanation for the correct option:

Given that, $9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)$

$\Rightarrow$ $225a^2+9b^2+25c^2-75ac=45ba+15bc$

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

$\Rightarrow$ ${(15a–3b)}^2+{(3b–5c)}^2+{(15a–5c)}^2=0$

It is possible when,

$15a–3b=0$, $3b–5c=0$and $15a–5c=0$

$\Rightarrow$$\left(15a=3b=5c\right)=k$

Hence, $a=\frac{k}{15},b=\frac{k}{3}$ and $c=\frac{k}{5}$

$\therefore b-a=\frac{4k}{15},c-a=\frac{2k}{15}$ and $b-c=\frac{2k}{15}$

$\Rightarrow$$c-a=b-c$

$\Rightarrow$ $2c=a+b$

$\therefore b,c,a$ are in A.P.

Hence, Option ‘A’ is Correct.