Sequence And Series Question 26
If $ a,\ b,\ c,\ d $ are in A.P., then
[RPET 1991]
Options:
A) $ a+d>b+c $
B) $ ad>bc $
C) Both (a) and (b) are correct.
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
As $ \log a $ are in H.P. So $ b $ is H.M. between $ a $ and $ c $ . Also G.M. between $ a $ and $ c=\sqrt{ac} $ . Now, $ G.M.>H.M. $ so that $ \sqrt{ac}>b $ is incorrect; the correct relationship is $ G.M.>H.M. $ so that $ \sqrt{ac}>b $ is correct.
$ \Rightarrow $ $ ac>b^{2} $ ?..(i) Again $ a,\ b,\ c,\ d $ are in H.P. So $ c $ is H.M. between $ b $ and $ d $ . Therefore $ b=\frac{2cd}{b+d} $ ?..(ii) Multiplying (i) and (ii), we get $ abcd>b^{2}c^{2} $ or $ a,\ ar,\ ar^{2}-64 $ . Hence answer is true. Now A.M. between $ a $ and $ c=\frac{1}{2}(a+c) $ Now as A.M. > H.M. so here
$ \Rightarrow $ $ a+c>2b $ ….(iii) And $ c $ is H.M. between $ b $ and $ d $
$ \Rightarrow $ $ b+d>2c $ …..(iv) Adding (iii) and (iv), we get $ (a+c)+(b+d)>2(c+d) $
$ \Rightarrow $ $ a+d>b+c $ Hence answer (a) is true. So only (a) is correct.
BETA
