Sequence And Series Question 633
Question: If $ a^{2}+ab^{2}+16c^{2}=2(3ab+6bc+4ac) $ , where $ a,b,c $ are non-zero numbers. Then $ a,b,c $ are in
[AMU 2005]
Options:
A) A.P
B) G.P
C) H.P
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ (a+2b+2c) $ $ (a-2b+2c)=a^{2}+4c^{2} $
Þ $ {{(a+2c)}^{2}}-{{(2b)}^{2}}=a^{2}+4c^{2} $
Þ $ a^{2}+4ac+4c^{2}-4b^{2}=a^{2}+4c^{2} $
Þ $ 4ac-4b^{2}=0 $
Þ $ b^{2}=ac $ Hence a, b, c are in G.P.
BETA
