JEE Main Solved Paper 2017 Question 7

Question: For any three positive real numbers a, b and c, $ 9(25a^{2}+b^{2})+25(c^{2}-3ac)=15b(3a+c). $ Then: JEE Main Solved Paper-2017

Options:

A) a, b and c are in G.P.

B) b, c and a are in G.P.

C) b, c and a are in A.P.

D) a, b and c are in A.P.

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ {{(15a)}^{2}}+{{(3b)}^{2}}+{{(5c)}^{2}}-(15a)(5c)-(15a)(3b) $ $ -(3b)(5c)=0 $ $ \frac{1}{2}[{{(15a-3b)}^{2}}+{{(3b-5c)}^{2}}+{{(5c-15a)}^{2}}]=0 $ it is possible when 15a = 3b = 5c
    $ \therefore $ $ b=\frac{5c}{3},,a=\frac{c}{3} $ $ a+b=2c $
    $ \Rightarrow $ $ b,c,,a $ in $ A.P. $