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

Answer: (B)

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

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

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

$225a^2+9b^2+25c^2-75ac-45ab-15bc=0$

$(15a{)}^2+(2b{)}^2+(5c{)}^2-(15a\left)\right(5c)-(15a\left)\right(3b)-(3b\left)\right(5c)=0$

$(15a{)}^2+(2b{)}^2+(5c{)}^2=(15a\left)\right(5c)+(15a\left)\right(3b)+(3b\left)\right(5c)$

$15a=3b=5c=k$

$b=5a$ and $c=3a$

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

$\therefore b-a=\frac{4k}{15}$

and $c-a=\frac{2k}{15}$

and $b-c=\frac{2k}{15}$

$\therefore c-a=b-c$

$\Rightarrow 2c=a+b$

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