A) \[a+d>b+c\]
B) \[ad>bc\]
C) Both (a) and (b)
D) None of these
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 \[\sqrt{ac}>b\]\[\Rightarrow \]\[ac>{{b}^{2}}\] ?..(i) Again \[a,\ b,\ c,\ d\] are in H.P. So \[c\] is H.M. between \[b\]and \[d\]. Therefore \[b=\frac{2ac}{a+b}\] ?..(ii) Multiplying (i) and (ii), we get \[abcd>{{b}^{2}}{{c}^{2}}\] or \[a,\ ar,\ a{{r}^{2}}-64\]. 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 \[(a+c)+(b+d)>2(b+c)\]\[\Rightarrow \]\[a+d>b+c\] Hence answer (a) is true. So both (a) and (b) are correct.
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