Sequence And Series Question 294
Question: Given $ a+d>b+c $ where $ a,\ b,\ c,\ d $ are real numbers, then
[Kurukshetra CEE 1998]
Options:
A) $ a,\ b,\ c,\ d $ are in A.P.
B) $ \frac{1}{a},\ \frac{1}{b},\ \frac{1}{c},\ \frac{1}{d} $ are in A.P.
C) $ (a+b),\ (b+c),\ (c+d),\ (a+d) $ are in A.P.
D) $ \frac{1}{a+b},\ \frac{1}{b+c},\ \frac{1}{c+d},\ \frac{1}{a+d} $ are in A.P.
Show Answer
Answer:
Correct Answer: B
Solution:
$ a+d>b+c $
$ \Rightarrow $ $ a+b+c+d>2b+2c $
$ \Rightarrow $ $ \frac{a+c}{2}+\frac{b+d}{2}>b+c $ $ \because $ $ \frac{a+c}{2}>b $ and $ \frac{b+d}{2}>c $ $ [\because \ A>H] $ $ b $ is the H.M. of $ a $ and $ c $ and their A.M. is $ \frac{a+c}{2} $ . $ c $ is H.M. of $ b $ and $ d $ and their A.M. is $ \frac{b+d}{2} $ . Hence, $ a,\ b,\ c,\ d $ are in H.P.
$ \Rightarrow $ $ \frac{1}{a},\ \frac{1}{b},\ \frac{1}{c},\ \frac{1}{d} $ are in A.P.
BETA
