Sequence And Series Question 316
Question: If $ a,,b,,c $ are in A.P. and $ a^{2},,b^{2},c^{2} $ are in H.P., then
[UPSEAT 2001]
Options:
A) $ a\ne b\ne c $
B) $ a^{2}=b^{2}=\frac{c^{2}}{2} $
C) $ a,,b,,c $ are in G.P.
D) $ \frac{-a}{2},b,c $ are in G.P
Show Answer
Answer:
Correct Answer: D
Solution:
a, b, c, are in A.P.
Þ 2b = a + c,b - a = c - b $ a^{2},b^{2},c^{2} $ are in H.P. $ \frac{1}{b^{2}}-\frac{1}{a^{2}}=\frac{1}{c^{2}}-\frac{1}{b^{2}} $
$ \Rightarrow \frac{a^{2}-b^{2}}{a^{2}b^{2}}=\frac{b^{2}-c^{2}}{b^{2}c^{2}} $
Þ $ (a-b)[c^{2}(a+b)-a^{2}(b+c)]=0 $ , $ [\because ,(b-c)=(a-b)] $
Þ $ a=b $ or $ c^{2}a+c^{2}b-a^{2}b-a^{2}c=0 $
Þ $ c^{2}a+c^{2}b-a^{2}b-a^{2}c=0 $
Þ $ ac,(c-a)=b,(a^{2}-c^{2}) $
Þ $ ac=-b,(c+a) $
Þ $ -ac=b.2b $
Þ $ b^{2}=(-a/2),c $ ,
$ \therefore -a/2,b,c $ are in G.P.
BETA
