Question

The correct option among the following is?

Answer 1

Answer is (1).

Explanation:

We can write `9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c)` as

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

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

or `(15a)^2+(3b)^2+(5c)^2_(15a)(5c)-(15a)(3b)-(3b)(5c)=0`

This is of type `u^2+v^2+w^2-uv-vw-uw=0`

and we can write it as `2u^2+2v^2+2w^2-2uv-2vw-2uw=0`

or `(u-v)^2+(v-w)^2+(u-w)^2=0`

As sum of three squares is `0`, each one of them is `0`

and hence `u=v=w` and in given case we have

`15a=3b=5c` and `b=5a` and `c=3a`

Hence our `a,b` and `c` are equivalent to `a,5a` and `3a`

and `a,c` and `b` - or `b,c` and `a` are in A.P.

Hence answer is (1).