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.
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.
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