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. $
