Let `a_(i) = i+ (1)/(i)` for i= 1 , 2 …………. , 20. Put
`p = (1)/(20) ( a_(1) + a_(2) +……… + a_(20))` and `q = (1)/(20) ((1)/(a_(1)) + (1)/(a_(2)) + ……… + (1)/(a_(20)))`. Then

Options

A. ` q in ( 0 , (22 - p)/(21))`
B. `q in ((22- p)/(21) , (2(22-p))/(21))`
C. `q in ((2 (22 - p))/(21), (22-p)/(7))`
D. `q in ((22 - p)/(7), (4 (22 -p))/(21))`

Answer

A

Explanation

`q gt 0` try to the contra prove that `q lt (22 - p)/(21)`
` q + (p)/(21) lt (22)/(21)`
`q + (p)/(21) = (1)/(20)[sum_(i = 1)^(20) (1)/(a_(i)) + (1)/(21) sum_(i = 1)^(20) a_(i)]`
`= (1)/(20)[sum_(i=1)^(20) ((i)/(i^(2) + 1)) + (1)/(21) sum_(i=1)^(20) ( i + (1)/(i))]`
`= (1)/(2) + (1)/(20) [sum_(i = 1)^(20) (i)/(i^(2) + 1) + sum_(i=1)^(20)((1)/(21i))]`
`(1)/(2) + (1)/(20)[(1)/(2) + sum_(i=2)^(20) (i)/(i^(2) + 1) + sum_(i=1)^(20) (1)/(21i)]`
`lt (1)/(2) + (1)/(20) [ (1)/(2) + (2)/(5) sum_(i=2)^(20) 1 + (1)/(21)sum_(i = 1)^(20)1]`
`lt (1)/(2) + (1)/(20) [ (1)/(2) + (2)/(5) xx 19 + (1)/(21) xx 20]`
`lt (1)/(2) + (1)/(20) (1 + 8 + 1)`
`lt (1)/(2) + (1)/(2)`
`lt (22)/(21)`