If a,b,c and d are four unequal positive numbers which are in A.P then

Options

A. `1/a+1/d gt 1/b+ 1/c`
B. `1/a+1/d lt 1/b+1/c`
C. `1/b+1/c gt 4/(a+d)`
D. `1/a+1/d=1/b+1/c`

Answer

A, C

Explanation

Let b=a+p,c=a+2p,d=a+3p (where p is common difference). Then,
`(1/a+1/d)/(1/b+1/c)=(1/a+1/(a+3p))/(1/(a+p)+1/(a+2p))`
`=((a+p)(a+2p))/(a+(a+3p))`
`=(a^(2)+3ap+2p^(2))/(a^(2)+3ap)gt1`
`therefore1/a+1/dgt1/b+1/c`
`(1/b+1/c)(a+d)=(1/(a+p)+1/(a+2p))(a+a+3p)`
`=((2a+3p)^(2))/(a^(2)+3ap+2p^(2))`
`=4+(p^(2))/(a^(2)+3ap+2p^(2))gt4`