Sequence And Series Question 326
Question: If $ a,\ b,\ c $ are in G.P. and $ \log a-\log 2b,\ \log 2b-\log 3c $ and $ \log 3c-\log a $ are in A.P., then $ a,\ b,\ c $ are the length of the sides of a triangle which is
Options:
A) Acute angled
B) Obtuse angled
C) Right angled
D) Equilateral
Show Answer
Answer:
Correct Answer: B
Solution:
As given $ b^{2}=ac $ and $ 2(\log 2b-\log 3c)=\log a-\log 2b+\log 3c-\log a $
$ \Rightarrow $ $ b^{2}=ac $ and $ 2b=3c $
$ \Rightarrow $ $ b=2a/3 $ and $ c=4a/9 $ Since $ a+b=\frac{5a}{3}>c,\ b+c=\frac{10a}{9},>a,\ c+a=\frac{13a}{9}>b $ It implies that $ a,\ b,\ c $ form $ a $ triangle with $ a $ as the greatest side. Now, let us find the greatest angle $ A $ of $ \Delta ABC $ by using the cosine formula. $ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=-\frac{29}{48}<0 $
$ \therefore $ The angle $ A $ is obtuse.
BETA
