Discussion Forum

For any three positive real numbers a,b and c. If 9(25a²+b²)+25(c²-3ac)  =15b (3a+c), then

Answer:

Explanation:

We have,

 225a²+9b²+25c²-75ac-45ab-15bc=0

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

$\Rightarrow \frac{1}{2}[(15a-3b)^{2}+(3b-5c)^{2}+(5c-15a)^{2}]=0$

$\Rightarrow 15a=3b,3b=5c$  and  $5c=15a$

$\therefore$         15a=3b=5c

$\Rightarrow  \frac{a}{1}=\frac{b}{5}=\frac{c}{3}=\lambda(say)$

$\Rightarrow  a=\lambda,b=5\lambda ,c=3\lambda$

   Hence,  a,b and c are in A.P